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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
int main() {
  int m = (int)1e6, ms[(int)1e6 + 1];
  for (int i = 0; i < m + 1; i++) ms[i] = 1;
  ms[0] = ms[1] = 0;
  for (int i = 2; i * i <= m; i++)
    if (ms[i])
      for (int j = i * i; j <= m; j += i) ms[j] = 0;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    long long a;
    scanf("%I64d", &a);
    int l = 2, r = (int)1e6, t;
    while (r - l > 1) {
      t = (r + l) >> 1;
      long long b = 1ll * t * t;
      if (b > a)
        r = t;
      else
        l = t;
    }
    if (1ll * l * l == a && ms[l])
      puts("YES");
    else if (1ll * r * r == a && ms[r])
      puts("YES");
    else
      puts("NO");
  }
}