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.
给你两个数a,b 都是 从1开始的 可以选择一个 *k 另一个*k^2， 问你a b分数可以达到吗，
因为任意一个数都是 *k 和*k*k 所以每次总增加都是 *k^3，所以总的结果其实相当于 i^3
输入a,b, 假设得出的结论由四个自然数形成， 则结果大致为 a= (k1^2) k2(k3^2)k4 和b= k1(k2^2)k3(k4^2) 这样的形式。 当两个数相乘结果为 (k1^3)(k2^3)(k3^3)*(k4^3)。 可以对该数求出其三次方根 temp = k1*k2*k3*k4. 则 有 x=a/temp= k1*k3, y=b/temp= k2*k3; 如果有x*x*y==a&&y*y*x==b 则可以判断答案是否正确。
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6+10;
map<long long,int> mp;
int main()
{
for(int i=1;i<N;i++)
mp[1LL*i*i*i]=i;
int t;
scanf("%d",&t);
cout<<cbrt(16*16)<<endl;
while(t--)
{
long long a,b;
scanf("%lld%lld",&a,&b);
if((mp[a*b]!=0)&&(a%mp[a*b]==0)&&(b%mp[a*b]==0)){//判断k1 k2 k3 k4
//是否存在
printf("Yes\n" );
}
else printf("No\n");
}
}