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 sang[1000010] = {0}, p[80000];
int n, m;
void taosang() {
  m = 0;
  for (int i = 2; i < 1000001; ++i)
    if (sang[i] == 0) {
      m++;
      p[m] = i;
      for (int j = i + i; j < 1000001; j += i) sang[j] = 1;
    }
}
bool isprime(long long x) {
  if (x < 2) return false;
  int sqrt_x = sqrt(x);
  for (int i = 3; i <= sqrt_x; ++i)
    if (x % i == 0) return false;
  return true;
}
int main() {
  taosang();
  long long x;
  int canx;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%I64d", &x);
    if (x == 1) {
      printf("NO\n");
      continue;
    }
    canx = sqrt(x);
    if (1LL * canx * canx != x)
      printf("NO\n");
    else if (sang[canx] == 0)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}