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;
int isPrime(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) {
      return 0;
    }
  }
  return 1;
}
int main() {
  long long n;
  cin >> n;
  int check;
  long long s;
  long long a[n];
  for (long long j = 0; j < n; j++) {
    cin >> a[j];
  }
  for (long long j = 0; j < n; j++) {
    s = sqrt(a[j]);
    check = isPrime(s);
    if (s * s == a[j] && s != 1) {
      if (check == 0) {
        cout << "NO" << endl;
      } else {
        cout << "YES" << endl;
      }
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}