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;
long long i, j;
long long isprime[1000000];
void prime() {
  isprime[0] = isprime[1] = 1;
  for (i = 2; i < 1000000; i++) {
    if (isprime[i] == 0) {
      for (j = i * i; j < 1000000; j += i) {
        isprime[j] = 1;
      }
    }
  }
}
int perfect_square(long long x) {
  double sqrt_x = sqrt(x);
  if (sqrt_x == int(sqrt_x)) {
    return 1;
  } else {
    return 0;
  }
}
int main() {
  prime();
  long long m, n;
  cin >> n;
  for (long long x = 0; x < n; x++) {
    cin >> m;
    if (m == 4) {
      cout << "YES" << endl;
    } else if (m % 2 == 0) {
      cout << "NO" << endl;
    } else if (perfect_square(m) == 1 && isprime[int(sqrt(m))] == 0) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}