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 int thieve10;
bool thieve3[1100003];
bool thieve2[1100003];
long long int thieve4[1100003];
void thieving() {
  long long int thieve = 2, thieve1;
  thieve3[0] = thieve3[1] = true;
  while (thieve < 1100003) {
    thieve4[thieve10] = thieve;
    thieve10++;
    for (thieve1 = thieve * 2; thieve1 < 1100003; thieve1 += thieve) {
      thieve3[thieve1] = true;
    }
    thieve++;
    while (thieve < 1100003 && thieve3[thieve]) thieve++;
  }
}
bool binarySearch(long long int array[97000], long int lowerbound,
                  long int upperbound, long long int key) {
  int position;
  position = (lowerbound + upperbound) / 2;
  while ((array[position] != key) && (lowerbound <= upperbound)) {
    if (array[position] > key) {
      upperbound = position - 1;
    } else {
      lowerbound = position + 1;
    }
    position = (lowerbound + upperbound) / 2;
  }
  if (lowerbound <= upperbound) {
    return true;
  } else
    return false;
}
int main() {
  thieving();
  long long int n, ni;
  cin >> ni;
  while (ni) {
    cin >> n;
    long int pol = int(sqrt(n));
    if (int(sqrt(n)) == sqrt(n) && (!thieve3[pol]))
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
    ni--;
  }
  return 0;
}