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 sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
bool prime[1000000 + 1];
void seive() {
  long long int N = 1e6 + 1;
  memset(prime, true, sizeof(prime));
  prime[0] = false;
  prime[1] = false;
  for (long long int i = 2; i * i <= N; i++) {
    if (prime[i]) {
      for (long long int j = i * i; j <= N; j += i) {
        prime[j] = false;
      }
    }
  }
}
bool isPowerOfTwo(int x) { return (x && !(x & (x - 1))); }
bool is_prime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long int prime_n(long long int x) {
  while (1) {
    int f = 1;
    for (long long int i = 2; i <= sqrt(x); i++) {
      if (x % i == 0) {
        f = 0;
        break;
      }
    }
    if (f)
      return x;
    else
      x++;
  }
}
void hemant() {
  long long int n;
  cin >> n;
  long long int x = sqrt(n);
  if (n == 1) {
    cout << "NO"
         << "\n";
    return;
  }
  if (x * x == n) {
    if (is_prime(x))
      cout << "YES"
           << "\n";
    else {
      cout << "NO"
           << "\n";
      return;
    }
  } else
    cout << "NO"
         << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  cin >> t;
  while (t--) {
    hemant();
  }
  return 0;
}