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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tt = 1;
  for (int tc = 1; tc <= tt; tc++) {
    int n;
    cin >> n;
    set<long long> primes;
    auto seive = [&]() {
      bool vis[1000001] = {};
      for (long long i = 2; i < 1000001; i++) {
        if (vis[i]) continue;
        primes.insert(i * i);
        vis[i] = true;
        for (int j = i * 2; j < 1000001; j += i) {
          vis[j] = true;
        }
      }
    };
    seive();
    for (int i = 0; i < n; i++) {
      long long a;
      cin >> a;
      if (*primes.lower_bound(a) == a)
        cout << "YES\n";
      else
        cout << "NO\n";
    }
  }
}