#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int cnt(int x){//�������� 
    int ans=0;
    for(int i=2;i*i<=x;i++)
        while(x%i==0){
          x/=i,ans++;
        }    
    if(x>1)ans++;
    return ans;
}
int main(){
    int t;
    cin>>t;
    while(t--){
    	int a,b,k;
        scanf("%d%d%d",&a,&b,&k);
        int mi;
        int ay=cnt(a),by=cnt(b);
        if(a==b)mi=0;
        else{
          if(a<b)swap(a,b);
          if(a%b==0)mi=1;
          else mi=2;
        }
    
        bool fl=false;//mi�����ٴ�����ay��by��a��b���������������� 
        if(k>=mi && k<=ay+by) fl=true;
        if(k==1 && mi!=1)fl=false; //��k=1ʱ��ֻ��mi=1��Ϊtrue 
        if(fl)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

You are given two integers a and b. In one turn, you can do one of the following operations: 

  * Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c; 
  * Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c. 



Your goal is to make a equal to b using exactly k turns.

For example, the numbers a=36 and b=48 can be made equal in 4 moves: 

  * c=6, divide b by c ⇒ a=36, b=8; 
  * c=2, divide a by c ⇒ a=18, b=8; 
  * c=9, divide a by c ⇒ a=2, b=8; 
  * c=4, divide b by c ⇒ a=2, b=2. 



For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).

Output

For each test case output: 

  * "Yes", if it is possible to make the numbers a and b equal in exactly k turns; 
  * "No" otherwise. 



The strings "Yes" and "No" can be output in any case.

Example

Input


8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1


Output


YES
YES
YES
YES
YES
NO
YES
NO

### Response

```cpp
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int cnt(int x){//�������� 
    int ans=0;
    for(int i=2;i*i<=x;i++)
        while(x%i==0){
          x/=i,ans++;
        }    
    if(x>1)ans++;
    return ans;
}
int main(){
    int t;
    cin>>t;
    while(t--){
    	int a,b,k;
        scanf("%d%d%d",&a,&b,&k);
        int mi;
        int ay=cnt(a),by=cnt(b);
        if(a==b)mi=0;
        else{
          if(a<b)swap(a,b);
          if(a%b==0)mi=1;
          else mi=2;
        }
    
        bool fl=false;//mi�����ٴ�����ay��by��a��b���������������� 
        if(k>=mi && k<=ay+by) fl=true;
        if(k==1 && mi!=1)fl=false; //��k=1ʱ��ֻ��mi=1��Ϊtrue 
        if(fl)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

```