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 int N = 1e6;
vector<bool> isprime(N + 1, true);
void seive() {
  isprime[0] = false;
  isprime[1] = false;
  for (int i = 2; i * i <= 1e6; i++) {
    if (isprime[i]) {
      for (int j = i * i; j <= 1e6; j += i) {
        isprime[j] = false;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  seive();
  long long int n;
  cin >> n;
  long long int a[n];
  long long int p, q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    p = sqrt(a[i]);
    q = ceil(sqrt(a[i]));
    if (p == q) {
      if (isprime[p])
        cout << "YES" << endl;
      else
        cout << "NO" << endl;
    } else
      cout << "NO" << endl;
  }
  return 0;
}