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 debug = false;
vector<int> G[(3 * 100000 + 42)];
long long phi(long long n, long long k) {
  long long result = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  vector<long long> V(1e6, true);
  for (long long i = 2; i < 1e6; i++) {
    int x = i * 2;
    while (x < 1e6) {
      V[x] = false;
      x += i;
    }
  }
  set<long long> P;
  for (long long i = 2; i < 1e6; i++)
    if (V[i]) P.insert(i * i);
  while (T--) {
    long long x;
    cin >> x;
    cout << (P.find(x) == P.end() ? "NO\n" : "YES\n");
  }
  return 0;
}