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;
const long long N = 1e6 + 100;
long long n, ehPrimo[N];
void sieve() {
  for (long long i = 0; i < N; i++) ehPrimo[i] = true;
  ehPrimo[2] = true;
  ehPrimo[1] = false;
  for (long long i = 2; i * i < N; i++) {
    if (ehPrimo[i]) {
      for (long long j = i + i; j < N; j += i) {
        ehPrimo[j] = false;
      }
    }
  }
}
int main() {
  sieve();
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) {
    long long x;
    scanf("%lld", &x);
    long long p = sqrt(x);
    if (p * p == x) {
      if (ehPrimo[p])
        printf("YES\n");
      else
        printf("NO\n");
    } else
      printf("NO\n");
  }
  return 0;
}