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 INF = 1e9 + 23;
const long long MOD = 1e9 + 7;
const int N = 1e6 + 10;
bool prime[N];
void pre() {
  fill(prime, prime + N, true);
  prime[1] = false;
  for (int i = (2); i < (N); ++i) {
    if (!prime[i]) continue;
    for (int j = (i + i); j < (N); j += (i)) prime[j] = false;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  pre();
  int n;
  cin >> n;
  for (int _ = 0; _ < (n); ++_) {
    long long x;
    cin >> x;
    long long sq = sqrt(x);
    if (sq * 1LL * sq != x || !prime[sq]) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
  }
  return 0;
}