CodeForces

Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.

The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k², and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.

Unfortunately, Slastyona had lost her notepad where the history of all n games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.

Input

In the first string, the number of games n (1 ≤ n ≤ 350000) is given.

Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 10⁹) – the results of Slastyona and Pushok, correspondingly.

Output

For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.

You can output each letter in arbitrary case (upper or lower).

Example

Input

62 475 458 816 16247 9941000000000 1000000

Output

YesYesYesNoNoYes

题目大概：

两个人说数字，谁的大谁赢，赢的一方的分数乘大的数字的平方，输的分数乘大的数字，先知道两方分数，问比赛评分是否对。（比赛可以有多轮）

思路：

当一轮比赛，两方分数相乘为大的数的三次方，当多轮时，也成立，为多轮大数乘积的三次方。用二分求出最后的值，检验该值是否对，并输出结果。

代码：

#include <iostream>#include <string>#include <queue>#include <cstring>#include <cstdio>using namespace std; long long go(long long sum){   long long l,r;   l=0;r=1000000;   while(l!=r)   {       long long mid=(l+r+1)/2;       if(mid*mid*mid>sum){       r=mid-1;       }       else {       l=mid;       }   }   return l;}int main(){int t;scanf("%d",&t);for(int i=0;i<t;i++){    long long n,m;    scanf("%I64d%I64d",&n,&m);  long long g=go(n*m); if(g*g*g!=n*m)printf("No\n");    else if(n%g==0&&m%g==0)printf("Yes\n");    else printf("No\n");} return 0;}