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() {
  long long int n, x, c = 0, j, a[100010];
  set<long long> s;
  bool vis[1000010];
  for (int i = 1; i < 1000001; ++i) vis[i] = true;
  for (int i = 2; i < 1000001; ++i) {
    if (vis[i] == true) {
      for (j = i * 2; j < 1000001; j += i) {
        vis[j] = false;
      }
    }
  }
  for (j = 2; j < 1000001; j++) {
    if (vis[j] == true) {
      s.insert(j * j);
    }
  }
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (s.find(a[i]) != s.end())
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}