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 isPrime(long long n) {
  if (n == 2 || n == 3 || n == 5) return true;
  if (n == 1 || n % 2 == 0 || n % 3 == 0 || n % 5 == 0) return false;
  long long sqrtN = (long long)sqrt(1.0 * n);
  for (long long i = 7, k = 4; i <= sqrtN; i += k, k = 6 - k)
    if (n % i == 0) return false;
  return true;
}
int main() {
  int n;
  vector<long long> a;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    a.push_back(x);
  }
  for (int i = 0; i < n; ++i) {
    long long x = (long long)sqrt(1.0 * a[i]);
    if (x == sqrt(1.0 * a[i]) && isPrime(x))
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}