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 a) {
  for (long long int i = 2; i * i <= a; i++)
    if (a % i == 0) return 0;
  return 1;
}
int main() {
  int T;
  scanf("%d", &T);
  for (int ii = 0; ii < T; ii++) {
    long long int n;
    scanf("%lld", &n);
    long long int k = (long long int)pow((double)n, 0.5);
    if (n == 1)
      printf("NO\n");
    else if (k * k == n && isprime(k) == 1)
      printf("YES\n");
    else if ((k + 1) * (k + 1) == n && isprime(k + 1) == 1)
      printf("YES\n");
    else
      printf("NO\n");
  }
}