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 a[1000007];
bool prime[10000007];
void isprime() {
  prime[0] = 0;
  prime[1] = 0;
  for (long long p = 2; p < 10000007; p++) {
    if (prime[p] == 1) {
      for (long long i = p * p; i < 10000007; i += p) prime[i] = 0;
    }
  }
}
int ps(long long n) {
  double x = sqrt(n);
  if (x == (int)x) {
    return 1;
  } else {
    return 0;
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  memset(prime, 1, sizeof(prime));
  isprime();
  long long t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> a[i];
    long long n = a[i];
    long long m = sqrt(n);
    if (n == 4)
      cout << "YES" << endl;
    else if (n % 2 == 0)
      cout << "NO" << endl;
    else if (ps(n) && prime[m])
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}