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;
int main() {
  const int mx = 1e6 + 1;
  int n;
  cin >> n;
  long long m, t;
  bool* a = new bool[n];
  bool* p = new bool[mx];
  p[0] = 0;
  p[1] = 0;
  p[2] = 1;
  for (int i = 3; i < mx; i++) p[i] = (i % 2);
  for (long long i = 3; i < mx; i += 2) {
    if (!p[i]) continue;
    for (long long j = i * i; j < mx; j += 2 * i) p[j] = 0;
  }
  for (int i = 0; i < n; i++) {
    a[i] = 0;
    cin >> m;
    t = sqrt(m);
    if (pow(t, 2) == m && p[t]) a[i] = 1;
  }
  for (int i = 0; i < n; i++) cout << (a[i] ? "YES" : "NO") << endl;
  return 0;
}