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

vector<int> factorize(int x)
{
    vector<int> v;
    for (int i = 2; i * i <= x; ++i)
    {
        while (x % i == 0) {
            v.push_back(i);
            x /= i;
        }
    }
    if (x != 1) {
        v.push_back(x);
    }
    return v;
}

signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);

    int T, a, b, k;
    cin >> T;
    while (T--)
    {
        cin >> a >> b >> k;
        if(k == 1)
        {
            int mx = max(a, b);
            int mn = min(a, b);
            if(a == b || mx % mn > 0)
                cout << "NO" << "\n";
            else
                cout << "YES" << "\n";
        }
        else
        {
            vector<int> V1 = factorize(a);
            vector<int> V2 = factorize(b);
            cout << (V1.size() + V2.size() >= k ? "YES\n" : "NO\n");
        }
    }

    return 0;
}
### Prompt

Please formulate 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;

vector<int> factorize(int x)
{
    vector<int> v;
    for (int i = 2; i * i <= x; ++i)
    {
        while (x % i == 0) {
            v.push_back(i);
            x /= i;
        }
    }
    if (x != 1) {
        v.push_back(x);
    }
    return v;
}

signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);

    int T, a, b, k;
    cin >> T;
    while (T--)
    {
        cin >> a >> b >> k;
        if(k == 1)
        {
            int mx = max(a, b);
            int mn = min(a, b);
            if(a == b || mx % mn > 0)
                cout << "NO" << "\n";
            else
                cout << "YES" << "\n";
        }
        else
        {
            vector<int> V1 = factorize(a);
            vector<int> V2 = factorize(b);
            cout << (V1.size() + V2.size() >= k ? "YES\n" : "NO\n");
        }
    }

    return 0;
}
```