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 = 100002;
long long n, n_0;
bool c[MAX];
vector<int> primes, fact;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 2; i < MAX; i++) {
    if (!c[i]) {
      for (int j = i + i; j < MAX; j += i) {
        c[j] = true;
      }
    }
  }
  while (t--) {
    primes.clear();
    fact.clear();
    cin >> n;
    n_0 = n;
    c[0] = c[1] = true;
    for (int i = 1; i <= sqrt(n) + 2; i++) {
      if (!c[i]) primes.push_back(i);
    }
    for (int i = 0; i < primes.size(); i++) {
      while (n % primes[i] == 0) {
        fact.push_back(primes[i]);
        n /= primes[i];
      }
    }
    if (n != 1) fact.push_back(n);
    sort(fact.begin(), fact.end());
    if (fact.size() < 3) {
      cout << "NO\n";
      continue;
    }
    long long a = fact[0], b = fact[1], c = fact[2];
    if (a == b) {
      b *= c;
      c = 1;
    }
    for (int i = 3; i < fact.size(); i++) {
      c *= fact[i];
    }
    if (a != c && b != c && a != b && (a * b * c) == n_0 && c != 1) {
      cout << "YES\n" << a << " " << b << " " << c << "\n";
    } else
      cout << "NO\n";
  }
}