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;
const int MAX_N = -1;
const int inf = 2e9 + 19;
vector<long long> vec;
void primeDiv(long long n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) vec.push_back(i), n /= i;
    }
  }
  if (n) vec.push_back(n);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long Q;
  cin >> Q;
  while (Q--) {
    long long t, a = 1, b = 1, c = 1;
    cin >> t;
    vec.clear();
    primeDiv(t);
    if ((int)vec.size() < 3) {
      cout << "NO" << endl;
      continue;
    }
    a = vec[0];
    b = vec[1];
    bool flag = false;
    if (a == b) b *= vec[2], flag = true;
    for (int i = 2 + flag; i < (int)vec.size(); i++) {
      c *= vec[i];
    }
    if (a < 2 || b < 2 || c < 2 || a == b || b == c || a == c)
      cout << "NO" << endl;
    else
      cout << "YES" << endl << a << ' ' << b << ' ' << c << endl;
  }
  return (0);
}