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 a = 1000000;
bool is_prime[1000005];
void sieve() {
  for (int i = 1; i <= a; i++) is_prime[i] = 1;
  is_prime[1] = 0;
  for (int i = 2; i * i <= a; i++) {
    if (is_prime[i]) {
      for (int j = 2; j * i <= a; j++) is_prime[i * j] = 0;
    }
  }
}
int main() {
  long long n = 0, i = 0, x = 0, y = 0, z = 0, j = 0, a[100005] = {0};
  cin >> n;
  sieve();
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) {
    x = a[i];
    y = sqrt(x);
    if (y * y != x) {
      cout << "NO\n";
      continue;
    }
    if (is_prime[y]) {
      cout << "YES\n";
      continue;
    } else {
      cout << "NO\n";
      continue;
    }
  }
  return 0;
}