#include <bits/stdc++.h>

using namespace std;

int cnt(int a) {
  int res = 0;
  for (int i = 2; i * i <= a; ++i) {
    while (a % i == 0) {
      a /= i;
      res++;
    }
  }
  if (a > 1) {
    res++;
  }
  return res;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  for (cin >> t; t--;) {
    int a, b, k;
    cin >> a >> b >> k;
    if (k == 1) {
      if ((a % b == 0 && a > b) || (b % a == 0 && b > a)) {
        cout << "YES\n";
      }
      else {
        cout << "NO\n";
      }
    }
    else {
      int u = cnt(a), v = cnt(b);
      if (2 <= k && k <= u + v) cout << "YES\n";
      else cout << "NO\n";
    }
  }
  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;

int cnt(int a) {
  int res = 0;
  for (int i = 2; i * i <= a; ++i) {
    while (a % i == 0) {
      a /= i;
      res++;
    }
  }
  if (a > 1) {
    res++;
  }
  return res;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  for (cin >> t; t--;) {
    int a, b, k;
    cin >> a >> b >> k;
    if (k == 1) {
      if ((a % b == 0 && a > b) || (b % a == 0 && b > a)) {
        cout << "YES\n";
      }
      else {
        cout << "NO\n";
      }
    }
    else {
      int u = cnt(a), v = cnt(b);
      if (2 <= k && k <= u + v) cout << "YES\n";
      else cout << "NO\n";
    }
  }
  return 0;
}

```