#include <bits/stdc++.h>
using namespace std;
#define fast ios_base :: sync_with_stdio(false),cin.tie(NULL);

int NoOfFactors(int x)
{
    int d=0;
    for(int i=2; i*i<=x; i++) {
        while(x%i==0)
        {
            d++;
            x/=i;
            if(x==1)break;
        }
    }
    if(x>1) d++;
    return d;
}

void solve(){
    int a, b, k;
    cin >> a >> b >> k;
    int fa = NoOfFactors(a);
    int fb = NoOfFactors(b);
    if (k == 1 && (a % b == 0 || b % a == 0) && (a != b)) {
        cout << "YES" << endl;
    }
    else if (k > 1 && k <=  fa + fb) {
        cout << "YES" << endl;
    }
    else {
        cout << "NO" << endl;
    }
}

int32_t main(){
    fast
    int t=1;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
### Prompt

Your challenge is to write a cpp solution 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;
#define fast ios_base :: sync_with_stdio(false),cin.tie(NULL);

int NoOfFactors(int x)
{
    int d=0;
    for(int i=2; i*i<=x; i++) {
        while(x%i==0)
        {
            d++;
            x/=i;
            if(x==1)break;
        }
    }
    if(x>1) d++;
    return d;
}

void solve(){
    int a, b, k;
    cin >> a >> b >> k;
    int fa = NoOfFactors(a);
    int fb = NoOfFactors(b);
    if (k == 1 && (a % b == 0 || b % a == 0) && (a != b)) {
        cout << "YES" << endl;
    }
    else if (k > 1 && k <=  fa + fb) {
        cout << "YES" << endl;
    }
    else {
        cout << "NO" << endl;
    }
}

int32_t main(){
    fast
    int t=1;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
```