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 is_simple(int num) noexcept {
  if (num < 2) return false;
  int bound = static_cast<int>(round(sqrt(num))) + 1;
  for (int i = 2; i < bound; ++i) {
    if (num % i == 0) return false;
  }
  return true;
}
int main() {
  int count = 0;
  std::cin >> count;
  for (int i = 0; i < count; ++i) {
    double cur_num;
    std::cin >> cur_num;
    double num = sqrt(cur_num);
    if (fabs(num - round(num)) < 0.00001) {
      if (is_simple(static_cast<int>(round(num))))
        std::cout << "YES\n";
      else
        std::cout << "NO\n";
    } else
      std::cout << "NO\n";
  }
}