//
// Created by asus on 2021/6/10.
//

//#include "D.h"
#include <bits/stdc++.h>

using namespace std;


vector<int> p;
int get(int x){
    int c = 0;
    for(int i : p){
        if(i>(x/i)) break;
        while(x%i==0) x/=i,c++;
    }
    if(x>1) c++;
    return c;
}

#define MX 31622

bool ok[MX+1];
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    for(int i = 2; i<=MX; i++){
        if(!ok[i]){
            p.push_back(i);
            for(int j = i; j<=MX; j+=i){
                ok[j] = 1;
            }
        }
    }
    int t;
    cin>>t;

    while(t--){
        int a,b,k;
        cin>>a>>b>>k;
        if(get(a)+get(b) >=k){
            if(k==1)
                cout << ((a!=b && (a%b==0 || b%a==0))?"YES\n":"NO\n");
            else
                cout<<"YES\n";

        }else cout<<"NO\n";
    }
}
### 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
//
// Created by asus on 2021/6/10.
//

//#include "D.h"
#include <bits/stdc++.h>

using namespace std;


vector<int> p;
int get(int x){
    int c = 0;
    for(int i : p){
        if(i>(x/i)) break;
        while(x%i==0) x/=i,c++;
    }
    if(x>1) c++;
    return c;
}

#define MX 31622

bool ok[MX+1];
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    for(int i = 2; i<=MX; i++){
        if(!ok[i]){
            p.push_back(i);
            for(int j = i; j<=MX; j+=i){
                ok[j] = 1;
            }
        }
    }
    int t;
    cin>>t;

    while(t--){
        int a,b,k;
        cin>>a>>b>>k;
        if(get(a)+get(b) >=k){
            if(k==1)
                cout << ((a!=b && (a%b==0 || b%a==0))?"YES\n":"NO\n");
            else
                cout<<"YES\n";

        }else cout<<"NO\n";
    }
}
```