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 long long INF = 1e9;
const long long N = 1e3 + 500;
const long long mod = 1e9 + 7;
const long double eps = 1E-7;
long long n, mx, mn, cnt, m, ans;
long long a, b, c;
string s;
long long used[N];
char z;
int main() {
  ios_base::sync_with_stdio(0);
  long long T;
  cin >> T;
  while (T--) {
    cin >> n;
    vector<long long> v;
    for (long long i = 2; i <= 1000000; i++) {
      if (n % i == 0 and v.size() < 2) {
        v.push_back(i);
        n /= i;
      } else if (v.size() == 2)
        break;
    }
    if (v.size() == 2 and n != v[0] and n != v[1] and n > 1) {
      v.push_back(n);
    }
    if (v.size() == 3)
      cout << "YES\n" << v[0] << " " << v[1] << " " << v[2] << endl;
    else
      cout << "NO\n";
    v.clear();
  }
}