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 isperfectsquare(long long n) {
  long long squareRootN = (long long)round((sqrt(n)));
  if (squareRootN * squareRootN == n) {
    return 1;
  } else {
    return 0;
  }
}
int isprime(int number) {
  if (number < 2) return 0;
  if (number == 2) return 1;
  if (number % 2 == 0) return 0;
  for (int i = 3; (i * i) <= number; i += 2) {
    if (number % i == 0) return 0;
  }
  return 1;
}
int main() {
  long long n, a[100000], i, j;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    long long t = a[i];
    if ((isperfectsquare(t) == 1) && (isprime(sqrt(t)) == 1))
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
}