#include <bits/stdc++.h>
using namespace std;
int f(int n)
{
    int cnt=0;
    while (n % 2 == 0)
    {
        cnt++;
        n = n/2;
    }
    for (int i = 3; i <= sqrt(n); i = i + 2)
    {
        while (n % i == 0)
        {
            cnt++;
            n = n/i;
        }
    }
    if (n > 2)
        cnt++;
    return cnt;
}
int main() {
    int t;
    cin>>t;
    while(t--)
    {
        int a,b,k;
        cin>>a>>b>>k;
        int mx=f(a)+f(b);
        if(k>=2 && k<=mx)
        {
            cout<<"YES"<<endl;
        }
        else if(k==1 && (a%b==0 || b%a==0) && a!=b)
        {
            cout<<"YES"<<endl;
        }
        else cout<<"NO"<<endl;
    }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 f(int n)
{
    int cnt=0;
    while (n % 2 == 0)
    {
        cnt++;
        n = n/2;
    }
    for (int i = 3; i <= sqrt(n); i = i + 2)
    {
        while (n % i == 0)
        {
            cnt++;
            n = n/i;
        }
    }
    if (n > 2)
        cnt++;
    return cnt;
}
int main() {
    int t;
    cin>>t;
    while(t--)
    {
        int a,b,k;
        cin>>a>>b>>k;
        int mx=f(a)+f(b);
        if(k>=2 && k<=mx)
        {
            cout<<"YES"<<endl;
        }
        else if(k==1 && (a%b==0 || b%a==0) && a!=b)
        {
            cout<<"YES"<<endl;
        }
        else cout<<"NO"<<endl;
    }
}

```