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 prime(long long n) {
  long long i, c = 1;
  for (i = 2; i <= sqrt(n); i += c) {
    if (n % i == 0) return 0;
    if (i == 3) c = 2;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (long long j = 0; j < n; j++) {
    long long x;
    cin >> x;
    if (x <= 2)
      cout << "NO"
           << "\n";
    else {
      long long u = sqrt(x);
      if (prime(u) && (u * u == x))
        cout << "YES"
             << "\n";
      else
        cout << "NO"
             << "\n";
    }
  }
  return 0;
}