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;
bool ans[10000007];
void isPrime(long long n) {
  memset(ans, true, sizeof(ans));
  ans[0] = 0;
  ans[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    for (long long j = i * 2; j <= n; j += i) {
      ans[j] = false;
    }
  }
}
int main() {
  isPrime(1000007);
  long long n;
  cin >> n;
  long long a[n], i;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < n; i++) {
    long long x = sqrt(a[i]);
    if ((x * x) != a[i] or ans[x] == false)
      cout << "NO" << endl;
    else
      cout << "YES" << endl;
  }
  return 0;
}