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;
using ll = long long;
using ld = long double;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
const int MOD = 1e9 + 7;
const int MAX = 1e5 + 5;
const ll INF = 1e18 + 1;
void err(...) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  for (auto i : *it) cerr << i;
  cerr << " = " << a << "\n";
  err(++it, args...);
}
template <class Ch, class Tr, class Container>
basic_ostream<Ch, Tr>& operator<<(basic_ostream<Ch, Tr>& os,
                                  Container const& x) {
  os << "{ ";
  for (auto& y : x) os << y << " ";
  return os << "}";
}
template <class X, class Y>
ostream& operator<<(ostream& os, pair<X, Y> const& p) {
  return os << "[" << p.first << " : " << p.second << "]";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    set<int> s;
    int a = -1;
    int b = -1;
    for (int i = 2; i < sqrt(n); i++) {
      if (n % i == 0 && n / i != i) {
        a = i;
        b = n / i;
        break;
      }
    }
    if (a == -1) {
      cout << "NO"
           << "\n";
      continue;
    }
    s.insert(a);
    s.insert(b);
    int c = -1, d = -1;
    for (int i = 2; i < sqrt(a); i++) {
      if (a % i == 0 && a / i != i && i != b && (a / i) != b) {
        c = i;
        d = a / i;
        break;
      }
    }
    if (c != -1) {
      cout << "YES"
           << "\n";
      cout << b << " " << c << " " << d << "\n";
      continue;
    }
    for (int i = 2; i < sqrt(b); i++) {
      if (b % i == 0 && b / i != i && i != a && (b / i) != a) {
        c = i;
        d = b / i;
        break;
      }
    }
    if (c != -1) {
      cout << "YES"
           << "\n";
      cout << a << " " << c << " " << d << "\n";
      continue;
    }
    cout << "NO"
         << "\n";
  }
  return 0;
}