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() {
  int n, size = 1e6 + 1;
  cin >> n;
  vector<int> prime(1e6 + 1, 1);
  prime[0] = 0;
  prime[1] = 0;
  for (int ri = 2; ri <= sqrt(size); ri++) {
    if (prime[ri]) {
      for (int pi = ri * ri; pi <= size; pi += ri)
        if (prime[pi]) prime[pi] = 0;
    }
  }
  vector<long long int> tprime;
  for (long long int ri = 0; ri < size; ri++) {
    if (prime[ri]) {
      tprime.push_back(ri * ri);
    }
  }
  long long int arr[n];
  for (int ri = 0; ri < n; ri++) scanf("%lld", &arr[ri]);
  for (int ri = 0; ri < n; ri++) {
    bool val = binary_search(tprime.begin(), tprime.end(), arr[ri]);
    if (val)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}