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>
#pragma GCC optimize("O2")
using namespace std;
const int INF = int(2e9) + 99;
void test_case() {
  long long n;
  cin >> n;
  set<long long> st;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0 && st.find(i) == st.end()) {
      st.insert(i);
      n /= i;
      break;
    }
  }
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0 && st.find(i) == st.end()) {
      st.insert(i);
      n /= i;
      break;
    }
  }
  if (st.size() < 2 || n == 1 || st.find(n) != st.end()) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  st.insert(n);
  for (auto x : st) cout << x << ' ';
  cout << '\n';
  return;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) test_case();
  return 0;
}