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 is_prime(long long x) {
  int flag = 1;
  long long f = sqrt(x);
  if (x == 2)
    return flag;
  else if (!(x % 2))
    flag = 0;
  else {
    for (long long i = 3; i <= f; i += 2) {
      if (x % i == 0) {
        flag = 0;
        break;
      }
    }
  }
  return flag;
}
int main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 0; i < n; i++) {
    long long a;
    scanf("%lld", &a);
    long long temp = sqrt(a);
    if ((a % 2 == 0 && a != 4) || a == 1) {
      printf("NO\n");
    } else {
      if (temp * temp == a) {
        int flag = is_prime(temp);
        if (flag)
          printf("YES\n");
        else
          printf("NO\n");
      } else {
        printf("NO\n");
      }
    }
  }
  return 0;
}