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;
const int N = 1e6;
bool prime[N + 10];
vector<long long> v;
void Sieve() {
  prime[0] = prime[1] = 1;
  for (long long i = 4; i <= N; i += 2) prime[i] = 1;
  for (long long i = 3; i <= N / i; i += 2) {
    if (!prime[i]) {
      for (long long j = i * i; j <= N; j += i + i) prime[j] = 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  Sieve();
  for (long long i = 2; i <= N; i += 1 + (i & 1)) {
    if (!prime[i]) {
      v.push_back(i * i);
    }
  }
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (binary_search(v.begin(), v.end(), n))
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}