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;
long long n, a;
bool b[1100000];
void f(long long x);
bool fp(int xo);
int main() {
  cin >> n;
  for (int i = 2; i <= 1000000; i++) {
    b[i] = fp(i);
  }
  for (int i = 0; i < n; i++) {
    cin >> a;
    f(a);
  }
  return 0;
}
void f(long long x) {
  long long xk = sqrt(x), ans = 0;
  if (xk * xk == x && xk != x) ans++;
  if (b[xk] == 1) {
    cout << "NO" << endl;
    return;
  }
  if (ans == 1)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
bool fp(int xo) {
  int xok = sqrt(xo);
  for (int i = 2; i <= xok; i++) {
    if (!(xo % i)) return 1;
  }
  return 0;
}