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;
pair<long long, int> a[100000];
bool ans[100000];
bool isprime(int m) {
  for (int i = 2; i * i <= m; i++)
    if (m % i == 0) return false;
  return true;
}
int main() {
  int n, i, j;
  while (1 == scanf("%d", &n)) {
    for (i = 0; i < n; i++) {
      scanf("%I64d", &a[i].first);
      a[i].second = i;
    }
    sort(a, a + n);
    long long nxt = 2;
    i = 0;
    while (i < n) {
      if (nxt * nxt == a[i].first) {
        ans[a[i].second] = true;
        i++;
      } else {
        if (a[i].first < nxt * nxt) {
          ans[a[i].second] = false;
          i++;
        } else if (a[i].first > nxt * nxt) {
          while (!isprime((int)++nxt))
            ;
        }
      }
    }
    for (i = 0; i < n; i++) puts(ans[i] ? "YES" : "NO");
  }
  return 0;
}