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;
vector<long long int> crivo;
vector<pair<long long int, long long int> > leitor;
bool bob[212345];
void fazCrivo() {
  long long int index = 0;
  for (long long int i = 2; i < 1000000; i++) {
    bool primo = true;
    if (i == 2)
      crivo.push_back(2);
    else {
      for (long long int j = 0;
           j < crivo.size() && crivo[j] < (long long int)(sqrt(i) + 1); j++) {
        if (i % crivo[j] == 0) {
          primo = false;
          break;
        }
      }
    }
    if (primo) {
      crivo.push_back(i);
      while (index < leitor.size() && leitor[index].first < i * i) {
        bob[leitor[index].second] = false;
        index++;
      }
      while (index < leitor.size() && leitor[index].first == i * i) {
        bob[leitor[index].second] = true;
        index++;
      }
      if (index >= leitor.size()) break;
    }
  }
}
int main() {
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    leitor.push_back(make_pair(x, i));
  }
  sort(leitor.begin(), leitor.end());
  fazCrivo();
  for (long long int i = 0; i < n; i++) {
    string aux = (bob[i]) ? "YES\n" : "NO\n";
    cout << aux;
  }
  return 0;
}