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;
long long n, m, ni, mi;
long long cn;
string s;
long long num[100005];
long long ok = false;
deque<long long> v;
map<string, bool> M;
map<string, bool> MC;
long long sch(long long a, long long b, long long c, long long i) {
  string s = to_string(a) + " " + to_string(b) + " " + to_string(c) + " " +
             to_string(i);
  if (M[s]) return MC[s];
  if (a > 1 && b > 1 && c > 1) {
    if (a != b && a != c && b != c) {
      if (a * b * c == cn) {
        cout << "YES" << endl;
        cout << a << " " << b << " " << c << endl;
        return 1;
      }
    }
  }
  long long na, nb, nc;
  if (i == v.size()) return 0;
  na = a * v[i];
  nb = b * v[i];
  nc = c * v[i];
  M[s] = 1;
  MC[s] = sch(na, b, c, i + 1) || sch(a, nb, c, i + 1) || sch(a, b, nc, i + 1);
  return MC[s];
}
void solve() {
  cin >> n;
  cn = n;
  v.clear();
  M.clear();
  MC.clear();
  for (long long i = 2; i * i < n + 5; i++) {
    if (n % i == 0) {
      while (n % i == 0) {
        v.push_back(i);
        n /= i;
      }
    }
  }
  if (n != 1) {
    v.push_back(n);
  }
  if (!sch(1, 1, 1, 0)) {
    cout << "NO" << endl;
  }
}
int32_t main() {
  cin.tie(0);
  ios_base::sync_with_stdio(NULL);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}