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>
bool xetso(unsigned long long n) {
  double c;
  c = sqrt(n);
  if (c == floor(c)) {
    if (c == 2 || c == 3) return true;
    if (c == 1 || (long long)c % 2 == 0 || (long long)c % 3 == 0) return false;
    for (long long i = 5; i <= sqrt(c); i += 6) {
      if ((long long)c % i == 0 || (long long)c % (i + 2) == 0) {
        return false;
      }
    }
    return true;
  } else
    return false;
}
int main() {
  unsigned long soso;
  unsigned long long so;
  std::cin >> soso;
  for (long i = 1; i <= soso; i++) {
    std::cin >> so;
    if (xetso(so))
      std::cout << "YES\n";
    else
      std::cout << "NO\n";
  }
}