#include<bits/stdc++.h>
using namespace std;
int t, a, b, k;
int get (int a) {
    int i = 2, res = 0;
    while (i * i <= a) {
        while (a % i == 0) {
            ++ res;
            a /= i;
        }
        ++ i;
        if (a == 0) break;
    }
    if (a > 1) ++ res;
    return res - 1;
}
void go () {
    cin >> a >> b >> k;
    if (k == 1) {
        if (a != b && (a % b == 0 || b % a == 0)) {
            cout << "YES\n";
        }
        else {
            cout << "NO\n";
        }
        return;
    }
    k -= 2;
    if (k < 0) {
        cout << "NO\n";
        return;
    }
    int tmp = get (a) + get (b);
    if (tmp >= k) {
        cout << "YES\n";
    }
    else {
        cout << "NO\n";
    }
    return;
}
int main () {
    ios_base::sync_with_stdio();
    cin.tie (0);
    cout.tie (0);
    cin >> t;
    while (t --) {
        go ();

    }
}

### Prompt

Create a solution in CPP for 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 t, a, b, k;
int get (int a) {
    int i = 2, res = 0;
    while (i * i <= a) {
        while (a % i == 0) {
            ++ res;
            a /= i;
        }
        ++ i;
        if (a == 0) break;
    }
    if (a > 1) ++ res;
    return res - 1;
}
void go () {
    cin >> a >> b >> k;
    if (k == 1) {
        if (a != b && (a % b == 0 || b % a == 0)) {
            cout << "YES\n";
        }
        else {
            cout << "NO\n";
        }
        return;
    }
    k -= 2;
    if (k < 0) {
        cout << "NO\n";
        return;
    }
    int tmp = get (a) + get (b);
    if (tmp >= k) {
        cout << "YES\n";
    }
    else {
        cout << "NO\n";
    }
    return;
}
int main () {
    ios_base::sync_with_stdio();
    cin.tie (0);
    cout.tie (0);
    cin >> t;
    while (t --) {
        go ();

    }
}

```