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;
long long i, j;
long long prime_flag[1000000];
void calculate_prime_flag() {
  prime_flag[0] = prime_flag[1] = 1;
  for (i = 2; i < 1000000; i++) {
    if (prime_flag[i] == 0) {
      for (j = i * i; j < 1000000; j += i) {
        prime_flag[j] = 1;
      }
    }
  }
}
int check_perfect_square(long long n) {
  double sqrt_n = sqrt(n);
  if (sqrt_n == int(sqrt_n)) {
    return 1;
  } else {
    return 0;
  }
}
int main() {
  calculate_prime_flag();
  long long total_numbers, n;
  cin >> total_numbers;
  for (i = 0; i < total_numbers; i++) {
    cin >> n;
    if (n == 4) {
      cout << "YES" << endl;
    } else if (n % 2 == 0) {
      cout << "NO" << endl;
    } else if (check_perfect_square(n) == 1 && prime_flag[int(sqrt(n))] == 0) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
}