#include<bits/stdc++.h>
using namespace std;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while(t--){
        int a, b, k;
        cin >> a >> b >> k;
        int refa = a, refb = b, fac = 0;
        for(int i = 2; i*i <= refa; i++){
            if(refa%i) continue;
            while(refa % i == 0){
                refa /= i;
                fac++;
            }
        }
        if(refa > 1) fac++;
        for(int i = 2; i*i <= refb; i++){
            if(refb%i) continue;
            while(refb % i == 0){
                refb /= i;
                fac++;
            }
        }
        if(refb > 1) fac++;
        if(a > b) swap(a, b);
        if(a == b){
            if(k == 0 || (k >= 2 && k <= fac)) cout << "Yes\n";
            else cout << "No\n";
        }
        else if(b % a == 0){
            if(k >= 1 && k <= fac) cout << "Yes\n";
            else cout << "No\n";
        }
        else if(k >= 2 && k <= fac) cout << "Yes\n";
        else cout << "No\n";
    }
    return 0;
}
### Prompt

Please create a solution in CPP to the following problem:

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<bits/stdc++.h>
using namespace std;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while(t--){
        int a, b, k;
        cin >> a >> b >> k;
        int refa = a, refb = b, fac = 0;
        for(int i = 2; i*i <= refa; i++){
            if(refa%i) continue;
            while(refa % i == 0){
                refa /= i;
                fac++;
            }
        }
        if(refa > 1) fac++;
        for(int i = 2; i*i <= refb; i++){
            if(refb%i) continue;
            while(refb % i == 0){
                refb /= i;
                fac++;
            }
        }
        if(refb > 1) fac++;
        if(a > b) swap(a, b);
        if(a == b){
            if(k == 0 || (k >= 2 && k <= fac)) cout << "Yes\n";
            else cout << "No\n";
        }
        else if(b % a == 0){
            if(k >= 1 && k <= fac) cout << "Yes\n";
            else cout << "No\n";
        }
        else if(k >= 2 && k <= fac) cout << "Yes\n";
        else cout << "No\n";
    }
    return 0;
}
```