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;
void sieve(long long int[]);
void findTprime(long long int[], long long int);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int size;
  long long int a[1000000], num;
  cin >> size;
  sieve(a);
  while (size--) {
    int cnt = 0;
    cin >> num;
    findTprime(a, num);
  }
  return 0;
}
void sieve(long long int a[]) {
  a[0] = a[1] = 0;
  for (int i = 2; i < 1000000; i++) a[i] = 1;
  for (int i = 2; i <= sqrt(1000000); i++) {
    if (a[i] == 1) {
      for (int j = 2; j * i <= 1000000; j++) {
        a[j * i] = 0;
      }
    }
  }
}
void findTprime(long long int a[], long long int num) {
  long long int flag = 0, temp;
  temp = sqrt(num);
  if ((long)temp * temp == num && a[temp] == 1) {
    flag = 1;
  }
  if (flag)
    cout << "YES\n";
  else
    cout << "NO\n";
}