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 int MX = 10e12 + 1;
template <typename T>
inline void read(T &x) {
  char c;
  bool nega = 0;
  while ((!isdigit(c = getchar())) && (c != '-'))
    ;
  if (c == '-') {
    nega = 1;
    c = getchar();
  }
  x = c - 48;
  while (isdigit(c = getchar())) x = x * 10 + c - 48;
  if (nega) x = -x;
}
void Solution() {
  int x;
  read(x);
  for (int i = 0; i < x; i++) {
    long long int s, y, j;
    read(s);
    y = sqrt(s);
    for (j = 2; j * j < y; j++)
      if (s % j == 0) break;
    printf("%s\n", (j * j > y and y * y == s and s > 1 ? "YES" : "NO"));
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  Solution();
  return 0;
}