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 int N = (int)1e6 + 1;
long long bin_search(double x) {
  double lo = 0, mid, hi = x;
  while (lo <= hi) {
    mid = (hi + lo) / 2;
    cout << mid << ' ';
    if (mid * mid == x)
      return (long long)mid;
    else if (mid * mid < x)
      lo = mid + 1;
    else
      hi = mid - 1;
  }
  return 0;
}
int main(int argc, char const *argv[]) {
  vector<bool> v(N, 0);
  v[0] = 1;
  v[1] = 1;
  for (long long i = 2; i < N; ++i)
    if (v[i] == 0)
      for (long long j = 2; i * j < N; j++) v[i * j] = 1;
  long long n, x, y;
  cin >> n;
  while (n--) {
    cin >> x;
    y = sqrt(x);
    if (y * y == x) {
      if (!v[y])
        cout << "YES\n";
      else
        cout << "NO\n";
    } else
      cout << "NO\n";
  }
  return 0;
}