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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long i, j, k, n;
    cin >> n;
    map<long long, long long> mp;
    long long x = n;
    while (x % 2 == 0) {
      mp[2]++;
      x /= 2;
    }
    for (i = 3; i * i <= x; i += 2) {
      while (x % i == 0) {
        mp[i]++;
        x /= i;
      }
    }
    if (x > 1) mp[x]++;
    long long a = 1, b = 1, c = 1;
    i = 0;
    auto it = mp.begin();
    a = it->first;
    if (it->second > 2) {
      b = it->first * it->first;
      c = n / (it->first * it->first * it->first);
    } else if (it->second == 2) {
      b = it->first;
      it++;
      if (it == mp.end()) {
        cout << "NO\n";
        continue;
      } else {
        b *= it->first;
        c = n / (a * b);
      }
    } else {
      it++;
      if (it == mp.end()) {
        cout << "NO\n";
        continue;
      } else {
        b *= it->first;
        c = n / (a * b);
      }
    }
    if (a == b || b == c || c == a || a < 2 || b < 2 || c < 2) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      cout << a << " " << b << " " << c << "\n";
    }
  }
}