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 prime[1000009];
long long isprime(long long x) {
  long long i = 2, flag = 1;
  if (x == 1) return 0;
  if (prime[x] != -1) return prime[x];
  while ((i * i) <= x) {
    if (x % i == 0) {
      flag = 0;
      break;
    }
    i = i + 1;
  }
  prime[x] = flag;
  return flag;
}
int main() {
  memset(prime, -1, sizeof(prime));
  long long n, i;
  cin >> n;
  long long a[n], flag[n];
  long double x, diff;
  flag[0] = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    x = sqrt(a[i]);
    diff = x - (long long)x;
    if (diff != 0.0) {
      flag[i] = 0;
      continue;
    } else {
      flag[i] = isprime((long long)x);
    }
  }
  for (i = 0; i < n; i++) {
    if (flag[i])
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}