Problem: We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer t Т-prime, if t has exactly three distinct positive divisors.

You are given an array of n positive integers. For each of them determine whether it is Т-prime or not.

Input

The first line contains a single positive integer, n (1 ≤ n ≤ 105), showing how many numbers are in the array. The next line contains n space-separated integers xi (1 ≤ xi ≤ 1012).

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.

Output

Print n lines: the i-th line should contain "YES" (without the quotes), if number xi is Т-prime, and "NO" (without the quotes), if it isn't.

Examples

Input

3
4 5 6


Output

YES
NO
NO

Note

The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int is_prime(long long int numero) {
  if (numero == 1 || (numero % 2 == 0 && numero > 2)) {
    return 0;
  }
  long long int raiz = sqrt(numero) + 1;
  for (long long int e = 3; e <= raiz; e += 2) {
    if (numero % e == 0) {
      return 0;
    }
  }
  return 1;
}
int t_prime(long long int numero) {
  long double raiz = sqrt(numero);
  long long int piso = floor(raiz);
  if (piso != raiz) {
    return 0;
  } else {
    if (is_prime(piso)) {
      return 1;
    } else {
      return 0;
    }
  }
}
int main() {
  int N;
  long long int q;
  int cont = 0;
  cin >> N;
  while (cont < N) {
    cont++;
    cin >> q;
    if (t_prime(q)) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}