Codeforces Round #426 (Div. 2)The Meaningless Game(思维+二分)

C. 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 theround. After that, the winner's score is multiplied byk², and the loser's score is multiplied byk. In the beginning of thegame, 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 scoresa,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

Note

First game might have been consisted of one round, in which the number2 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 number5, and in the second one, Pushok would have barked the number3.

思路：：既然两个数里每次操作，都会有一个数 * x  另一个数 *x*x   那么两个数的积必定是某个数三次幂，就算是多次，最终累积也得是某个数三次幂。因此只要二分，看是否能找到某个数的三次幂 == 两数和就行了 ，但是存在三次幂只是必要条件，例如 1,  8,就是反例，还要加上  a*a % b == 0 && b*b%a == 0。保证是三次幂 2 ：1 或 1 ： 2 分给两个人.

#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <map>#include <algorithm>#include <iostream>using namespace std;typedef long long ll;bool judge(ll x, ll y){    ll tmp = x*y;    ll l = 1 , r = 1e6;    while(l<r)    {        ll mid = (l+r) / 2;        if(mid*mid*mid >= tmp)            r = mid;        else            l = mid+1;    }    return r*r*r == tmp && x*x%y==0 &&  y*y%x==0;}int main(){    int n;    scanf("%d",&n);    while(n--)    {        ll x,y;        scanf("%lld%lld",&x,&y);        printf(judge(x,y)?"YES\n":"NO\n");    }    return 0;}