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;
int N;
long long v;
bool crivo[1000024];
int eh(long long v) {
  long long lb = 1ll, ub = 1000000ll, m;
  while (lb != ub) {
    m = (lb + ub + 1ll) >> 1ll;
    if (m * m > v) {
      ub = m - 1;
    } else {
      lb = m;
    }
  }
  return (lb * lb == v && !crivo[lb]);
}
int main() {
  int i, n;
  memset(crivo, 0, sizeof(crivo));
  crivo[0] = 1;
  crivo[1] = 1;
  for (i = 2; i <= 1000000; i++) {
    if (crivo[i]) {
      continue;
    }
    for (n = i + i; n <= 1000000; n += i) {
      crivo[n] = 1;
    }
  }
  scanf("%d", &N);
  while (N--) {
    scanf("%I64d", &v);
    printf(eh(v) ? "YES\n" : "NO\n");
  }
  return 0;
}