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 isPerfectSquare(long long n) {
  long long squareRootN = (long long)round((sqrt(n)));
  if (squareRootN * squareRootN == n) {
    return true;
  } else {
    return false;
  }
}
bool isPrime(long num) {
  if ((int)num < 2)
    return false;
  else if (num == 2)
    return true;
  else if ((int)num % 2 == 0)
    return false;
  else {
    bool prime = true;
    int divisor = 3;
    double num_d = static_cast<double>(num);
    int upperLimit = static_cast<int>(sqrt(num_d) + 1);
    while (divisor <= upperLimit) {
      if ((int)num % divisor == 0) prime = false;
      divisor += 2;
    }
    return prime;
  }
}
int main() {
  int n;
  cin >> n;
  long long arr[n];
  bool result[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  for (int i = 0, k = (n / 2); k < n; k++) {
    if (i < n / 2) {
      if (isPerfectSquare(arr[i])) {
        if (isPrime((long)sqrt(arr[i])))
          result[i] = true;
        else
          result[i] = false;
      } else
        result[i] = false;
      i++;
    }
    if (isPerfectSquare(arr[k])) {
      if (isPrime((long)sqrt(arr[k])))
        result[k] = true;
      else
        result[k] = false;
    } else
      result[k] = false;
  }
  for (int i = 0; i < n; i++) {
    if (result[i])
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
}