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 sieve[1000002];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long xd = 1000000;
  vector<long long> vec;
  for (unsigned long long i = 2; i <= xd; i++) {
    if (!sieve[i]) {
      vec.push_back(i);
      for (long long j = 2 * i; j <= 1000000; j += i) sieve[j] = 1;
    }
  }
  set<long long> tprim;
  for (unsigned long long i = 0; i < vec.size(); i++) {
    tprim.insert(vec[i] * vec[i]);
  }
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    if (tprim.count(x))
      cout << "YES";
    else
      cout << "NO";
    cout << '\n';
  }
  return 0;
}