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;
template <class T>
void printArray(vector<T> arr) {
  for (T a : arr) cout << a << " ";
  cout << '\n';
}
void printVerdict(bool verdict) { cout << (verdict ? "YES" : "NO") << '\n'; }
vector<long long> findPrime(long long n) {
  vector<long long> ret;
  for (int i = 2; i * i <= n && ret.empty(); i++) {
    if (n % i == 0) ret.push_back(i);
  }
  if (ret.empty()) return {};
  for (int i = 2; i * i <= n / ret.front(); i++) {
    if ((n / ret.front()) % i == 0 && i != ret.front()) {
      ret.push_back(i);
      break;
    }
  }
  ret.push_back((n / ret.front()) / ret.back());
  if (ret.back() == ret[0] || ret.back() == ret[1]) return {};
  return ret;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    bool verdict = true;
    vector<long long> ret = findPrime(n);
    if (ret.empty()) verdict = false;
    printVerdict(verdict);
    if (verdict) printArray(ret);
  }
}