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 vi = vector<long long int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vc = vector<char>;
using vs = vector<string>;
using vld = vector<long double>;
using pii = pair<long long int, long long int>;
using psi = pair<string, long long int>;
using pci = pair<char, long long int>;
using vpii = vector<pii>;
long long int mod = 1e9 + 7;
long long int const maxn = 1e5 + 5;
long long int const inf = 1e18;
long long int add(long long int a, long long int b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long int mul(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int powm(long long int x, long long int n, long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long int modinverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
bool prime(long long int x) {
  if (x <= 1) return false;
  for (int i = 2; i <= sqrt(x); i++)
    if (x % i == 0) return false;
  return true;
}
long long int divisor(long long int x) {
  long long int cnt = 0;
  for (int i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      if (i != x / i)
        cnt += 2;
      else
        cnt += 1;
    }
  }
  return cnt;
}
vector<long long int> sieve(long long int n) {
  bool prim[n + 1];
  memset(prim, true, sizeof(prim));
  for (long long int p = 2; p * p <= n; p++) {
    if (prim[p] == true) {
      for (int i = p * p; i <= n; i += p) prim[i] = false;
    }
  }
  vector<long long int> v;
  for (int i = 2; i <= n; i++)
    if (prim[i]) v.push_back(i);
  return v;
}
void solve() {
  set<long long int> st;
  long long int x = 0;
  long long int n;
  cin >> n;
  long long int m = n;
  for (int i = 2; i * i <= m and x < 2; i++) {
    if (n % i == 0) {
      st.insert(i);
      x++;
      n /= i;
    }
  }
  st.insert(n);
  if ((long long int)st.size() == 3) {
    cout << "YES\n";
    for (auto it = st.begin(); it != st.end(); it++) cout << *it << " ";
    cout << endl;
  } else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}