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 ara[200005];
char st[200004];
long long aa[200004];
long long bb[200004];
long long isprime(long long n) {
  long long i, j;
  if (n == 2)
    return 1;
  else if (n % 2 == 0)
    return 0;
  else {
    j = sqrt(n) + 1;
    for (i = 3; i <= j; i += 2) {
      if (n % i == 0) return 0;
    }
    return 1;
  }
}
int main() {
  long long a, b, c, d, m, n, i, j, k, l, t, x, y, z;
  cin >> t;
  while (t--) {
    long long ans = 0;
    long long maxi = 0, mini = 0, sum = 0, mul = 1;
    cin >> n;
    a = sqrt(n);
    if (n == 4)
      cout << "YES" << endl;
    else if (a * a == n && n != 1 && n % 2 != 0 && isprime(a))
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}