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;
long long i, N, j, check = 1;
long long test, sq;
long long n[1000000];
int main() {
  cin >> N;
  for (i = 3; i <= 1000; i = i + 2) {
    if (n[i] == 0) {
      for (j = i * i; j <= 1000000; j += i + i) n[j] = 1;
    }
  }
  for (i = 0; i < N; i++) {
    cin >> test;
    sq = sqrt(test);
    if (sq * sq == test && test != 1) {
      if (test == 4)
        cout << "YES" << endl;
      else if (test % 2 == 0)
        cout << "NO" << endl;
      else if (n[sq] == 0)
        cout << "YES" << endl;
      else
        cout << "NO" << endl;
    } else
      cout << "NO" << endl;
  }
  return 0;
}