V The Meaningless Game---思维题

题目

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.

题解

题目大意

主人和狗做游戏，初始积分为(1)，每次选定一个整数k，谁赢了积分就乘(k^2)，输了就乘(k)，给出各自的积分，判断有没有可能出现这样的情况。

解题思路

这道题其实很简单 每次都这么说
我们这样想，每次都说一个数乘(k)，另一个乘(k^2)，那每次两数的积就乘(k^3)，最后两数的积就是一个数的三次方。
这里还要注意一种情况，就是其中一个数就达到了三次方，这样显然是不成立的，需要特判一下
还需要注意数据范围，两个1e9的数想乘会爆int，所以要用long long ;

代码

#include <cstdio>
#include <cmath>
#define int long long
using namespace std;
int n, x, y, k;
signed main() {
    scanf("%lld", &n);
    while (n--) {
        scanf("%lld%lld", &x, &y);
        k = pow((double)(x * y), 1.0 / 3) + 0.5;
        // + 0.5 是四舍五入
        if (k * k * k != x * y || x % k || y % k) puts("No");
        else puts("Yes");
    }
    return 0;
}