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>
int isprime(long long int n) {
  long long int i;
  if (n < 2) return 0;
  if (n == 2) return 1;
  for (i = 3; i * i <= n; i += 2) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int main() {
  long long int n, m, i, j;
  long long int x = 0, c = 0;
  scanf("%I64d", &n);
  for (i = 1; i <= n; i++) {
    c = 0;
    scanf("%I64d", &m);
    if (m % 2 != 0 && m != 1) {
      x = sqrt(m);
      if (x * x == m && isprime(x)) {
        c = 1;
        printf("YES\n");
        continue;
      } else {
        printf("NO\n");
        continue;
      }
    } else if (m == 4)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}