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;
template <typename tn>
inline tn next(void) {
  tn k;
  cin >> k;
  return k;
}
template <typename tn>
inline ostream& operator<<(ostream& os, const vector<tn>& v) {
  for (unsigned i = 0; i < v.size(); i++) os << v[i] << ' ';
  return os;
}
inline bool isPrime(int e) {
  if (e <= 1) return 0;
  for (int i = 2; i <= sqrt(e); i++)
    if (!(e % i)) return 0;
  return 1;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T = next<int>();
  while (T--) {
    long long a = next<long long>();
    double s = sqrt(a);
    if (s == (int)s && isPrime(s))
      puts("YES");
    else
      puts("NO");
  }
  return ~~(0 - 0);
}