833A The Meaningless Game

A. The Meaningless Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000

output

Yes
Yes
Yes
No
No
Yes

Note

First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.

The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.

这一题看得出技巧的会觉得很简单，但像我这样看不出技巧的就。。。。

如题意：喊的快的乘k的平方，喊得慢的乘k；

所以a*b就是k的三次方的倍数。

即只需判断a*b是否三次方数，开三次方后，是不是a和b的因子即可。

代码如下：

#include <cstdio>
#include <cmath>
int main(){
    long long  n;
    long long a,b;
    scanf("%lld",&n);
    while(n--){
        scanf("%lld %lld",&a,&b);
        long long u=floor(pow(a*b,1.0/3.0)+0.5);
        if(u*u*u==a*b&&!(a%u)&&!(b%u)) puts("Yes");
        else puts("No");
    }
    return 0;
}