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 int hcf(long long int a, long long int b) {
  if (b == 0) return a;
  return hcf(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / hcf(a, b);
}
bool isprime(long long int n) {
  if (n == 1) return false;
  for (long long int i = 2; i * i <= n; ++i) {
    if (n % i == 0) return false;
  }
  return true;
}
vector<long long int> primeFactors(long long int n) {
  vector<long long int> ans;
  while (n % 2 == 0) {
    ans.push_back(2);
    n = n / 2;
  }
  for (long long int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      ans.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) ans.push_back(n);
  return ans;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a = primeFactors(n);
  sort(a.begin(), a.end());
  if (a.size() == 3 and a[0] != a[1] and a[1] != a[2]) {
    cout << "YES"
         << "\n"
         << a[0] << " " << a[1] << " " << a[2] << "\n";
    return;
  }
  if (a.size() > 3 and a.front() != a.back()) {
    cout << "YES"
         << "\n"
         << a[0] << " ";
    long long int temp = 1;
    for (long long int i = 1; i <= a.size() - 2; i++) temp *= a[i];
    cout << temp << " " << a.back() << "\n";
    return;
  }
  if (a.size() > 5) {
    cout << "YES "
         << "\n"
         << a[0] << " ";
    long long int temp = 1;
    long long int temp2 = 1;
    for (long long int i = 1; i <= a.size() - 3; i++) temp *= a[i];
    for (long long int i = a.size() - 2; i <= a.size() - 1; i++) temp2 *= a[i];
    cout << temp << " " << temp2 << "\n";
    return;
  }
  {
    cout << "NO"
         << "\n";
    return;
  }
}
void onlinejudge() {}
int main() {
  onlinejudge();
  long long int ttt;
  ttt = 1;
  cin >> ttt;
  cout << fixed << setprecision(16);
  while (ttt--) solve();
  return 0;
}