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;
static long long int r, j, a, b, l, x, y, n, s, q, i, z, p, k, c, d, t;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> t;
  long long int a[1000001] = {0};
  vector<int> v;
  for (i = 2; i <= 1000000; i++) {
    if (a[i] == 0) {
      for (j = 2 * i; j <= 1000000; j = j + i) a[j] = 1;
    }
  }
  for (int i = 2; i < 1000001; i++) {
    if (a[i] == 0) {
      v.push_back(i);
    }
  }
  for (int j = 0; j < t; j++) {
    cin >> n;
    s = n;
    n = sqrt(n);
    if (n * n == s) {
      if (binary_search(v.begin(), v.end(), n))
        cout << "YES";
      else
        cout << "NO";
    } else
      cout << "NO";
    cout << endl;
  }
}