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;
bool isPrime(long long n) {
  int skip = 0;
  if (n < 2)
    return false;
  else if (n == 2)
    return true;
  long long limit = sqrt(n);
  if (n % 2 == 0) return false;
  for (int j = 3; j <= limit; j += 2) {
    if (n % j == 0) return false;
  }
  return true;
}
int main() {
  long long num;
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> num;
    long long sqr = sqrt(num);
    if (sqr * sqr == num && isPrime(sqr) == true)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}