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 long long mod = 1000000000, mod2 = 2e9;
double Pi = 3.141592653589793238;
unsigned long long gcd(unsigned long long a, unsigned long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
unsigned long long FP(unsigned long long b, unsigned long long p) {
  if (p == 0) return 1;
  return (p % 2 == 1) ? (b)*FP(b * b, p / 2) : FP(b * b, p / 2);
}
int arr[1000001] = {};
void solve() {
  for (long long i = 2; i < 1000001; i++) {
    if (arr[i] == 0) {
      long long j = i + i;
      while (j < 1000001) {
        arr[j] = 1;
        j += i;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  vector<long long> ans;
  solve();
  for (long long i = 2; i < 1000001; i++) {
    if (arr[i] == 0) ans.push_back(i * i);
  }
  long long x, t;
  cin >> t;
  for (int i = (0); i < (t); i++) {
    cin >> x;
    if (binary_search(ans.begin(), ans.end(), x))
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}