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;
bool TP[1000001] = {false};
vector<long long> P;
long long N, X, T[100000];
void primos() {
  P.push_back(4);
  for (long long i = 3; i < 1000000; i += 2) {
    if (!TP[i]) {
      P.push_back(i * i);
      TP[i] = false;
      for (long long j = 1; j * i <= 1000001; j++) {
        TP[j * i] = true;
      }
    }
  }
}
int main() {
  cin >> N;
  primos();
  for (int i = 0; i < N; i++) {
    cin >> X;
    T[i] = X;
  }
  for (long long i = 0; i < N; i++) {
    printf("%s\n", binary_search(P.begin(), P.end(), T[i]) ? "YES" : "NO");
  }
  return 0;
}