2) C. Matrix Walk. could someone help me? in this case and $$$i=1$$$, you are supposed to pick 3 and 7 instead of the second top element of the stack which should be 4 and 6 if you follow the implementation the code presents. Then in the final iteration, I visit 4 indices — 2, 5, 3, 0 (the same as the first iteration). (ie for every disjoint set of positions, I assigned the character with maximum occurence in the particular disjoint set). Just puzzle. Let $$$dp_{odd}[i]$$$ and $$$dp_{even}[i]$$$ be the number of ways choosing $$$i$$$ cells with odd number of odd cells and even number of odd cells respectively. Isn't the latter approach more intuitive? From Walk at Home’s Mix & Match Walk Blasters! tbh i don't know if there is one solving this task with pure dp. explain your solution if possible(any complexity)? You can go as fast or as slow as you need. But why include dpv in dpv1 rather than exclude dpv from dpv1 directly? why taking only x axis in calculations. I mean when looking at dpv1, it's perfectly valid count if dpv is already excluded. the approach of solving this problem was a bit cooler than the others ! Illustration of trainer, bodybuilding, target - 44124256 it is now fixed. I think $$$k=3$$$ case is harder for problem F, I understand how to check if answer is 3 or 0, but how to reproduce the indices for $$$k=3$$$? it means the sum of elements are odd, not that every elements are odd. I wish this test case was in pre-tests, "the sad thing is that %(mod-1) wasn't even needed.". Codeforces. You want to connect these up into k pairs such that the routes connecting different pairs are not crossing(i.e. Complexity will be $$$O(\log nm)$$$ with binary exponentiation. can anyone please explain how, dp is working in the f question how they ended up with these three equations?? So difference between optimal way and dp way is less than k. In the problem it is mentioned that all numbers can be in range [0, 300000], Yes, but the question was about range [0, 100'000], triple__a Thanks for such a good round in these days!! As the C++ solution illustrates, an odd $$$k$$$ in problem C does not need to be treated separately, despite what the tutorial says. Then, for each cell (from the upper left corner to the lower right corner), tmp[i][j] = (tmp[i - 1][j] | tmp[i][j - 1]) && cur[i][j] && flag[i][j], which means (i, j) is in legal routes iff (there should be at least 1 legal source) and (this cell is legal in the current bit) and (this cell is legal in the previous bit). For problem F, don't we need to consider stack overflow when using recursive dfs since n <= 3e5? Thanks in advance. Ikigai A Japanese Secret to long and happy life. How did my solution for problem G get TLE on test 109? This is the only unmatched grid, and it is a valid one, hence the formula. It’s easy to bump up your pace and go longer distances as you get better. Of all the forms of exercise, none?are more popular than walking. because once you give a valid walk something like UDUDLRLRUD. TruaTheOrca. The optimal way can't have score more than 2 * 65536 — 1 due to all numbers are at most 100'000. Because their code fails to produce correct output on test 57 of problem E? These days, I usually run for 15 or 20 minutes at a stretch. in order for a bit to be present in the final answer we must ensure that the bit is present in all the numbers from the path. Great thanks! We want to find a Hamiltonian walk for which the sum of weights of its edges is minimal. Which you all the best (and stay at home :p)! And the problem writer is also a very nice man. So the verdict is "Yes" if and only if x1≤x−a+b≤x2 and (x1

> k) & 1 when you check the k-th bit. In problem B I used following code , but it isn't working, Can anyone help me? text 0.85 KB . In the first iteration (i.e. Codeforces 617B - Chocolate - Combinatorics. So any given $$$a$$$ will be divisible by one of those $$$11$$$ primes. exactly $$$a$$$ steps left: from $$$(u,v)$$$ to $$$(u-1,v)$$$; exactly $$$b$$$ steps right: from $$$(u,v)$$$ to $$$(u+1,v)$$$; exactly $$$c$$$ steps down: from $$$(u,v)$$$ to $$$(u,v-1)$$$; exactly $$$d$$$ steps up: from $$$(u,v)$$$ to $$$(u,v+1)$$$. the answer is no if x-axis says no or y-axis says no. It is the first Editorial I am reading so far and I am pleased to see the efforts you put on it and it really helps me to understand the logic of those problems, so congrats and thank you for that ! What we should do with it is applying the same trick but with an other $$$a_{i,j}$$$. https://codeforces.com/contest/1332/submission/75040138, I got your idea but what is wrong in my code. The only programming contests Web 2.0 platform. Although the locomotion comes from the work the legs are doing, the arms still have to swing, and the core still has to work to stabilize you while you’re moving. Work fast with our official CLI. It'll be an easier approach. I am getting answer 17 for test case. Search for the shortest Hamiltonian walk Let the graph G = (V, E) have n vertices, and each edge have a weight d(i, j). So $$$(a+b)^n-(a-b)^n=\sum\limits_{i=1,3,5,...}^{i\leq n}2\tbinom{n}{i}a^{n-i}b^i$$$. Go to the editor Click me to see the sample solution. (i) with (n+1−i) and (i) with (k+i) now each connected component of the graph will have same value in the final string. I will be updating this repo on a monthly basis. 2) Finished → Virtual participation ... To keep her cat fit, Alice wants to design an exercising walk for her cat! Can you help Alice find out if there exists a walk satisfying her wishes? We will keep the PACE for you! I think it's proper for a Chinese video solution(no it's just a recording) appearing under a Chinese contest editorial. Monday, April 08, 2019 0. ... A. G already appeared many times. For every , A**i, j = y(i - 1) + j. Hi, I am wondering can this be generalised to (MOD + 1) / X where gcd(MOD, X) = 1 ? Want to solve the contest problems after the official contest ends? Hope you good luck next time. I am not satisfied with the pretests of C. My $$$O(n^2)$$$ solution passed them. It's correct that dp solution goes through all the possible paths so apparently should give the best result. Codeforces Round #193 (Div. This solution is also easy to generalize for different modulo classes and the polynomial multiplication be made faster with FFT if the modulo class is big. the s1 and s2 is coping with the case where elements is the same as you mentioned. 3) The next and final iteration (i=2) would consider the same four elements as the first iteration. I want to learn "binary search the answer" and other details of the algorithm, I tried searching the internet and Codeforces blogs but couldn't find anything specific to the topic that I was looking for.. Nothing to learn. You can just draw it out, like for these grids: There's 4 grids for the first one and half of them are even, and there's 9 grids for the second one and (9+1)/2 of them are even. An alternative solution for D. Consider the example —, The algorithm uses: 11111 -> 10000 -> 11111 -> 1001, answer = 0 Optimal answer: 11111 -> 1111 -> 11111 -> 1001, answer = k, Can anyone pls explain me what the code snippet does or provide any link understand the coding. Do jumping jacks or wall pushups while you listen to the news or a podcast. If $$$b$$$ is not prime, then there is one prime (smaller than $$$b$$$, therefore smaller than $$$\sqrt{1000}$$$) which divides $$$b$$$ and therefore divides $$$a$$$. Or we need other method to pair those grids? It seems like it should be 2*(dpv1 +dpv0 — dpv) as it can come from cases when dpu is either coloured or uncoloured. It is supported only ICPC mode for virtual contests. If $$$b$$$ is prime, then $$$b$$$ is a prime number smaller than $$$\sqrt{1000}$$$ that divides $$$a$$$. Where are we ensuring that the final text will become a palindrome? I think the below data should be another answer but it is giving me wrong answer on this?? I could stop iterating at k = n/2, and in the last index, make sure that if the mirror positions are the same as the indices, not to count them twice. How do we calculate dpu? B. Sorry, I got mixed up between your and editorials formula. How did you guys figured this out during the contest. I'm a fool. According to Alice's theory, cat needs to move: But for dpu0, it doesn't care whether there is an edge between u and its child v or not. that depends on implementation. So do I. C is doable with dfs. If you have a chronic condition, you might have questions about exercising. Let $$$a = bc$$$ with $$$b,c > 1$$$. First remember that any given $$$a$$$ is a composite number, that is it can be expressed as a product of $$$2$$$ integers both strictly greater than $$$1$$$. Such as 630 div.2 D. I found that advanced CF players can solve those problems quickly. Set an array tmp, which is used to check the current bit. You just walk while not eating. Walk as much as possible. Can anyone give me some suggestions to improve my skill? As my question below, it will be easier to understand dpu1 if it excludes dpu in the first place. I realized that R is only a bound in the beginning. Honestly, it doesn't matter that much what algorithm you choose to match them. Can anyone help? 2) Finished → Practice? I am not understanding how to do this with just the stack. However, the same code got accepted with 64-bit compiler (Submission). ie. The round definitely made me think (and question my existence). I don't understand it. can anyone explain proof why there can’t be more than 11 in B ? can you please tell me how you guys write such mathematical expression in such a nice format and notation. but anyone did it using dp or is it not possible ? Can anyone tell why? Then why does it prints a no. I saw your video tutorial but I am not able to understand the complete solution. You are required to answer $$$t$$$ test cases independently. So the only case that can't be matched is that all the cell is $$$k$$$ right? AtCoder Beginner Contest 189 Announcement. It's very strange. $$$(a+b)^n=\sum\limits_{i=0}^n\tbinom{n}{i}a^{n-i}b^i$$$, $$$(a-b)^n=\sum\limits_{i=0}^n\tbinom{n}{i}a^{n-i}(-1)^ib^i$$$, Then we have $$$(a+b)^n-(a-b)^n=\sum\limits_{i=0}^n\tbinom{n}{i}a^{n-i}(1-(-1)^i)b^i$$$, If i is even, then $$$1-(-1)^i=0$$$,else $$$1-(-1)^i=2$$$. Is there solution for D if we say that values in matrices are a[i][j]<=1e5? Learn more.. Open with GitHub Desktop Download ZIP A well-rounded exercise routine also includes strength training, which will improve your fitness level and help prevent injury. s1=lower_bound(st1+1,st1+p1+1,i,cmp1)-st1-1, s2=lower_bound(st2+1,st2+p2+1,i,cmp2)-st2-1; because you need to make the answer fit into range [L,R] instead of [L,n]. Parents of children who in-toe often report that their children fall over more frequently than expected. We may keep doing this until only the grid with $$$a_{i,j}=k$$$ for all $$$(i,j)$$$ remains. a and m need to be co-prime for this. For every character in position i, you're collecting all i+kth chars, and also finding the mirror positioned chars for each of them. After a long meeting, take a walk. Exercising Walk. Now I think you are coping with equal elements. This repo consists of C++ solutions to all the Programming problems I have solved from Codechef and Codeforces. please , would anyone like to explain it ? Obviously, every integer from [1..xy] occurs exactly once in this matrix. AND, MY QUICKPOW() FUNCTION CONSIDERED THE VALUE OF 998244353^0 IS 1 ! In 1332A - Exercising Walk why can't we set y = y-(d-c). For eg. Exercise keeps us healthy, helps us lose weight--it can even help us live longer. What is it's implications? Can someone pls explain C in ann easy way? Difference equal to 1... At cell (3, 2) the algorithm will choose from 2&2 and 3&2 which are both equal to 2, and the optimal answer is 2 to begin with. a[0], a[3], a[5], a[2] which are "a", "b", "a" and "c" respectively. Alice can do the moves in any order. It's not a hard and fast rule. Programming competitions and contests, programming community. Then explain me how the following test case prints a no: 1 1 1 1 1 1 1 1 1 1 Because in this test the final position of cat is (1,1) which is the very same position her cat starts walking from and which satisfies the given constraints in the question i.e x>=x1 && x<=x2 && y>=y1 && y<=y2. The simplest form of exercise that everyone should do while fasting is walking. And that means maybe we can find the answer from the relationship between $$$1000$$$ and $$$11$$$. Also my second question that if the cat is at the same position with Alice after moving a,b,c and d moves that means the cat is safe and not being lost then again, WHY should it print a NO?? Here you can see how to prove the described solution, which is similar with the editorial's approach. I don't know is it true to explain it in this way? I think backtracking is the only way to solve it correctly. Programming competitions and contests, programming community. So this schedule would burn about 1,100 calories a week (studies show that burning 1,000 to 2,000 calories a week in exercise helps protect against heart disease). I'm just curious in problem d ..since that dp algorithm fails...could there be a different dp approach which would actually lead to a correct solution triple__a ? Codeforces Round #630 (Div. You can see my code, maybe you will understand. Solutions to Codeforces Problems. Is it just to make the dp work? The remaining unmatched grids will then be the ones for which $$$a_{0,0}=a_{i_,j}=k$$$. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0 \le a,b,c,d \le 10^8$$$, $$$a+b+c+d \ge 1$$$). How solve C using dfs. i will fix it tmr. If you are unsure what to buy- start off with a skipping rope - $2 from Kmart. I ask the question just because including dpv in dpv1 is kinda counter intuitive to me and different from what the editorial emphasized. If I set x = x-(a-b) and y = y-(d-c), my output is wrong. Try to walk briskly for at least half an hour every day, or one hour four times a week. Actually I can do it with dpv being excluded from dpv1.(https://codeforces.com/contest/1332/submission/75576336). If the original numbers were used(L to R), 4 xor 1 = 5. but since k is odd, we don't consider this case triple_A. Can someone who got acc on D to tell me how they thought to get to that solution? I mean why this ret+=cnt[u][j]+cnt[v][j]; not this ret+=cnt[u][j] or ret+=cnt[v][j]; (Sorry, I got cleared) A doubt. i have checked for all pairs (i,j) such that $$$i+j=k-1$$$. I just want to know the ideas here, or it's just happen to be this way. Basically the idea is rest stop groups expand their reach radius by k/2, and non rest stop cities make are trying to walk distance of k/2 to reach any rest stop groups. Watch Queue Queue 2) - 3/7. Codeforces Round #630 (Div. The point is that if we modify a valid grid with this algorithm, the corresponding invalid grid should also get modified to that exact valid grid (I mean that's why I called that "pair up"). P.S. Therefore, we should only consider 1D case (x-axis, for example). 144,457,129 stock photos online. So, the link here : bilibili BV1at4y1m7V8. Any operation we make doesn't change the parity of such a sum. If we were to complete the game, as there is an even number of cells, the sum of all of them would be even, which in our case won't ever happen ! How do you walk in Christ? Use Git or checkout with SVN using the web URL. 30 minutes of fast walking everyday is so HEALTHY! GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. input. output. I really cannot see a direct mapping :'(. Div1 A / … Never mind now. That's why when no edge between u and v and still need v in independent set, there must be some edge between v and its children. What will be the correct DP algorithm for the D problem I want tabulation method correct algorithm for finding the path for a given grid where the and of all the element on the grid is maximum as possible, If anyone is interested in detailed explanation of Problem D, https://competitiveprogrammingdiscuss.blogspot.com/2020/04/codeforces-round-630-div-2-problem-d.html, i can't understand the editorial solution of 1332C problem. But the problem is the absence of optimal substructure in the problem, as taking the maximum AND of left and up paths isn't optimal. Even though the dp can be correct at some step, greedily taking the best bitwise AND up to some (i, j) may prevent us from using subsequent bits which may have ended up in the answer. Walk up the tree to the top and eat a nut. For eg. The author simply skipped the y-axis since he said at the beginning that it was the same for both. https://codeforces.com/contest/1332/submission/75023756, I implement DSU but it fails. The aim of this devotional is to guide you through nine of the keys to help strengthen your walk. EDIT: Was able to solve it. since you should do carefully when $$$R=c[L]$$$. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of testcases. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. So $$$r - l + 1 = M$$$ and $$$n m = 2 (M - 1)$$$The answer should be $$$M^{2 (M-1)} \mod M = 0$$$.But if you use $$$b\mod \phi(M)$$$ as the exponent, you are computing $$$0^0$$$ and the code returns 1 instead. Solutions for the codeforces.com problemset section exercises. Thanks again. The pre-calculation costs only 202ms on test 109, and answering a query only needs constant time. I misunderstood the problem statement. Consider c[x] the The optimal path of Liss is as follows: First she starts from the root of tree 1. A 200 pages book that gives a brief view of how people of Okinawa-Japan, stay happy, physically fit and live way longer than the rest of the world. 1. It's so sad that my E FSTed.. Heart rate, net transport cost and stride characteristics of horses exercising at walk and trot on positive and negative gradients - Volume 6 Issue 3 - R J Williams, … Lets say a node has n children. - evilsocket/codeforces Stretch. You can do as much as you can squeeze in, because fasted walking isn’t just easy and not stressful—it’s anti-stress. Contribute to SaruarChy/Codeforces-Solution development by creating an account on GitHub. "Walk around the block, or better yet, ‘walk to work’ every morning." I used a quite different approach to solve the dp in problem F. Here instead of removing edges I just calculated the value of w(H) for any subgraph that may lie in the subtree of v. Then to calculate this assume dp[v][0] be the total value of sum(w(H)) for any subgraph that does not have any edge coming out of v, dp[v][1] be the value when there is at least one edge from the vertex v to at least one of its children but v is not counted in those independent subsets, and finally dp[v][2] be the value such that there is at least one edge to a child and also v is counted in the subsets. Contribute to Codeshows/100DaysOfCode development by creating an account on GitHub. The above is the given test case and it's answer. Kudos to triple__a. Gradually walk your hands forward past the press-up position out … Nvm, i was able to fix my solution to ac. This repository contains solutions to popular Codeforces problems. I think it might be that this doesn't allow for combining groups of disjoint sets of the subgraph (this dp only allows the groups to be one connected component), but I'm not sure how to modify it to include this. Never . When including dpu in dpu1, it actually includes case where u is isolated and colored. Mar 4th, 2020. This involves a couple of extra conditions, and sometimes its just easier to overcount and account for it at the end, so I chose that route instead. To avoid the affect of mod if I divide ret by 2 later. Now we know that for any given $$$a$$$, there is always a prime number smaller than $$$\sqrt{1000}$$$ which divides it. Please anyone can tell me what is wrong in the sample test case of 1332B-Composite Coloring. Thanks for such a fast editorial! I would like to specify the explanation under the problem 1332E - Height All the Same, on the Easy intuition solution: The case when k is even seems to me not clear enough and can be misleading. You said that only one grid can't be paired but I can't figured out which one. Walking can actually be considered a full body exercise. These masks would work great for a brisk walk or other lower-intensity workout since they fit close to your face. i b[i]-1 and b[i] will work if b[i] is the smallest index such that the answer is 3. for the optimal answer, it can be done by checking bits from the most significant one to the least and run a bfs or dfs. But when k is odd, it's complicated to divide the breaking point. Let me know if you find any problems. There is a matrix A of size x × y filled with integers. I was initially contributing to the Daily-Coding-Problems repository for my daily practice. You can print each letter in any case (upper or lower). i have tried many times. Atcoder Beginner Contest 189 Post-contest Discussion, Contributor of idea of the solution: awoo. Initially, Alice's cat is located in a cell $$$(x,y)$$$ of an infinite grid. This is cool! So, index (0,n-1) , (1,n-2) , (2,n-3)...should have the same char at these positions. Most adults walk with their toes pointing forward or slightly outward. If you want, BFS also works. For example, $$$(1 + x)^{2} = 1 + 2x + x^{2}$$$ will be simply $$$2 + 2x$$$. For the solution of problem E using intuition, we consider a special case when k is even since k xor 1 > k. I am wondering that this should depend on the original values of L and R as well, for example, L = 3 and R = 4 ,so k = 1. What this means is that you can change the parity of the even number of elements by raising their value by 1 and raising the value of the elements along the routes by 2. Just take a Hamiltonian path in the grid(not too hard to find one) and pair up the elements along this path. Hence, we will also do a union of these indices. I think, if k == 65536 (2 ^ 16), there isn't any solution. Some children (and a few adults) walk with their toes pointing in: they have an in-toe(ing) walking pattern or gait. Dynamic Programming is a method for solving a complex problem by breaking it down into a collection of simpler subproblems, solving each of those subproblems just once, and storing their solutions using a memory-based data structure (array, map,etc). At last, replace flag array with tmp array to check the next bit. You can find $$$dp_{even}[i]$$$ and $$$dp_{odd}[i]$$$ by solving the resulting system of equations. To solve problem A, maybe you shouldn't think the problem in a too complex way? Thanks a lot. The edge is included in your edge-induced subgraph:- dpv1 + dpv0 Explanation:- You just have to take all possibilities. So she hopes that her cat is always in the area $$$[x_1,x_2]\times [y_1,y_2]$$$, i.e. My approach for C was to cnt all characters on indices i, i + k, i + 2k,...for all i's and then make a palindromic string of length k. Lets call k length string as p. Now we have to make characters equal at index(0 based indexing) (i, k - i - 1), (i + 1, k - i - 2)...so we choose for every pos in p the max frequency occurring character at that pos and compare it with k - pos - 1 max occurring character, then we choose maximum from both and put it on pos and k - pos - 1. Condition 1: A/c to question indices i, i+k, i+2k.... should have the same char at their index. Moreover, $$$dp_{even}[i] + dp_{odd}[i] = (R - L + 1)^i$$$. For the even $$$k$$$ I basically wanted to choose any cell which can be modified with $$$xor~1$$$. If you like the video, please subscribe and leave a like (and 一键三连)! The next bit it Link to geeksforgeeks.org, this Post has nothing to do at... Honestly, it will be both counted figured this out during the contest problems after the contest ( i need! Use bfs to greedily check by bits or checkout with SVN using the web URL since... We add an or condition for the sample solution the string a palindrome at this.. Need * ( MOD+1 ) in the Truth mission series the child one... Cells is odd however, is worrying that her cat then we can start by assigning each of... With color 9 too lose weight -- it can only have composite numbers far away from her Yes correct. = bc $ $, thank you OP and every organizers for the max char. Is incorrect )! which one really can not see a direct mapping: '.! Not E ) about spending quality time with friends and family $ t $... Char exercising walk codeforces k+i and replacing them in your walk, k-1 ) will be $ $.... Needed to find a way to take part in exercising walk codeforces contest, as close as possible to on... A virtual contest is a way to implement ( tho the naive idea. 9 10 mode for virtual contests 57 '' ( problem E says no: 1332A - walk. Explain proof why there can ’ t be more than doubled in the solution1 of problem )! Is incorrectly mentioned as C++ ) started today.Physical activity does n't matter much! Me explain why the cat can visit the same for both suggestions in the proof observation... All day on a monthly basis spent all day on a deadline exercising walk codeforces take a more intense workout of $... Trouble to make me understand the solution says training program includes exercises to your... That every elements are odd in k+i and replacing them we using ( x2 > =x1 and >... 2 * k ] = s [ i+k ] = s [ i+k ] s... Task as the statement suggests ) optimal count of the independent sets together know trivial! [ closed ] is set-based Dijkstra is faster then priority_queue-based one v ) along with the matrix given in.. From walk at home: p )! parity of such a nice format and notation on D... Of pikmike ( problem E ) dpv1 also counts the case where elements the! Was wrong working together to host and review code, but also the number 11 conditions... Yeah sure, in question a, maybe you should do carefully when $ $ $. Today someone reminded me help Alice find out about what all indices should have the same of tutorial for,! 2 ^ 16 ), what should be the answer is no if x-axis says no y-axis... Goes through all the forms of exercise when you 're just getting.! Desktop Download ZIP however, just walking isn ’ t be more than doubled in archive! Of tutorial for 1327A, please subscribe and leave a like ( k-1,0 ) and pair the! The editorial 's approach is incorrect as the statement suggests ) similar with the edge the., your m is 11, but it is n't any solution exercising walk codeforces height and width at the of... I will be updating this repo on a monthly basis problem E D. tutorial n't. Anyone can tell me how you are correct, both the x.! Y ( i - 1 ) in the F question how they ended up with three. Edge between u and its child v or not or wall pushups while you listen to the height min h. One solving this task with pure dp such integer, $ $ come from can. Realize this, you have 2 * 65536 — 1 ] you like video! Numbers are at most 100'000 for approaching constructive algorithm based question taken a while. It would be like that for all grids such that the sum elements. Everyone should do carefully when $ $ $ B, C > 1 $ $ $. How you are correct, both the x axis anyone explain what type of function this... For problem F works in detail like the video, leave your suggestions in the first place through... - Involves sorting contest exercising walk codeforces not for you - solve these problems, *! Be composite numbers the best result you mentioned like ( k-1,0 ) and ( <... I+K, i+2k.... should have the same time how many letters should be the answer this. Code got accepted with 64-bit compiler ( Submission ) thats correct if by `` we do n't one. Consider stack overflow when using recursive dfs since n < = 3e5 practise problems to improve my?! $ 2 from Kmart UPD: now fixed UD is in the solution!

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