The Meaningless Game
time limit per test 1 second memory limit per test 256 megabytes
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.

传送门:CF833A

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).

Sample Input

1
2
3
4
5
6
7
6
2 4
75 45
8 8
16 16
247 994
1000000000 1000000

Sample Output

1
2
3
4
5
6
Yes
Yes
Yes
No
No
Yes

题解

其实挺简单的 首先乘起来开三次方(得到加了多少次(需判断整数)) 然后判断2个条件 1:立方根是个整数 2:a和b能够整除立方根(肯定得是加的次数的倍数)
## AC code:(不包含输入类)
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import java.io.*;  
import java.util.*;
public class Main {
static double eps=1e-6;
public static void main(String[] args) {
FastScanner sc = new FastScanner();
PrintWriter pw = new PrintWriter(System.out);
while (sc.hasNext()) {
int n=sc.nextInt();
for(int o=1;o<=n;o++){
long a=sc.nextLong();
long b=sc.nextLong();
double c=Math.pow(a*b,1.0/3);
c=Math.round(c);
boolean flag=false;
if(Math.abs(c*c*c-a*b)<eps){
if(a%c==0&&b%c==0)flag=true;
}
if(flag)pw.println("Yes");
else pw.println("No");
}
pw.flush();
}
}
}