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 <typename T, typename TT>
ostream& operator<<(ostream& os, const pair<T, TT>& t) {
  return os << t.first << " " << t.second;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& t) {
  for (auto& i : t) os << i << " ";
  return os;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  scanf("%lld", &t);
  while (t--) {
    long long n;
    scanf("%lld", &n);
    vector<long long> v;
    long long tmp = n;
    for (int i = 2; i * i <= tmp; i++) {
      while (tmp % i == 0) {
        v.push_back(i);
        tmp /= i;
      }
    }
    if (tmp > 1) v.push_back(tmp);
    if (v.size() < 3) {
      cout << "NO" << '\n';
      continue;
    }
    long long a = -1, b = -1, c = -1;
    a = v[0];
    b = v[1];
    if (b == a) b *= v[2];
    c = n / (a * b);
    if (c == b or c == a or c <= 1)
      cout << "NO" << '\n';
    else {
      cout << "YES" << '\n' << a << " " << b << " " << c << '\n';
    }
  }
  return 0;
}