#include<bits/stdc++.h>
 
using namespace std;
 
int count(int n) {
	int k = 0;
	while (n % 2 == 0) {
		k++;
		n /= 2;
	}
	for (int p = 3; p <= n / p; p += 2)
		while (n % p == 0) {
			k++;
			n /= p;
		}
	if (n > 1)
		k++;
	return k;
}
 
int main() {
	int t; cin >> t;
	while (t--) {
		int a, b, k; cin >> a >> b >> k;
		if (k == 1) {
			cout << (a != b && (a % b == 0 || b % a == 0) ? "YES" : "NO") << '\n';
			continue;
		}
		int x = count(a);
		int y = count(b);
		if (k > x + y) {
			cout << "NO\n";
			continue;
		}
		cout << "YES\n";
	}
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 count(int n) {
	int k = 0;
	while (n % 2 == 0) {
		k++;
		n /= 2;
	}
	for (int p = 3; p <= n / p; p += 2)
		while (n % p == 0) {
			k++;
			n /= p;
		}
	if (n > 1)
		k++;
	return k;
}
 
int main() {
	int t; cin >> t;
	while (t--) {
		int a, b, k; cin >> a >> b >> k;
		if (k == 1) {
			cout << (a != b && (a % b == 0 || b % a == 0) ? "YES" : "NO") << '\n';
			continue;
		}
		int x = count(a);
		int y = count(b);
		if (k > x + y) {
			cout << "NO\n";
			continue;
		}
		cout << "YES\n";
	}
	return 0;
}
```