Problem: You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.

If there are several answers, you can print any.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).

Output

For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.

Otherwise, print "YES" and any possible such representation.

Example

Input


5
64
32
97
2
12345


Output


YES
2 4 8 
NO
NO
NO
YES
3 5 823 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline void setPrecision(int n) { cout.precision(n); }
long long int INF = 1e10;
int MOD = 1e9 + 7;
inline bool same2(int x, int y, int z) {
  return (x == y) || (y == z) || (z == x);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    int first = -1;
    for (int i = 2; i * i <= n; ++i) {
      if (n % i == 0) {
        first = i;
        break;
      }
    }
    if (first == -1) {
      cout << "NO\n";
      continue;
    }
    n /= first;
    int second = -1;
    for (int i = first + 1; i * i <= n; ++i) {
      if (n % i == 0) {
        second = i;
        break;
      }
    }
    if (second == -1 || same2(first, second, n / second)) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    cout << first << " " << second << " " << n / second << "\n";
  }
  return 0;
}