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;
using std::vector;
int isPrime(unsigned long long int n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
int t_prime(vector<long long int> &num) {
  for (int i = 0; i < num.size(); i++) {
    long long int temp = num[i];
    long long int start = 0;
    long long int end = num[i] / 2;
    int result = 0;
    if (temp == 0 || temp == 1) result = 0;
    if (temp > 4 && temp % 2 == 0)
      result = 0;
    else {
      while (start <= end) {
        long long int mid = (start + end) / 2;
        unsigned long long int prod = mid * mid;
        if (prod == temp) {
          if (isPrime(mid) == 1) result = 1;
          break;
        }
        if (prod < temp) {
          start = mid + 1;
        } else {
          end = mid - 1;
        }
      }
    }
    if (result == 1)
      cout << "YES"
           << "\n";
    else
      cout << "NO"
           << "\n";
  }
  return 0;
}
int main() {
  int n;
  cin >> n;
  vector<long long int> num(n);
  for (size_t i = 0; i < num.size(); ++i) {
    cin >> num[i];
  }
  int result;
  t_prime(num);
  return 0;
}