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 int maxn = 100007;
long long a[maxn];
const long long MAXN = 1000007;
int prime[MAXN];
bool vis[MAXN];
void Prime(long long n) {
  int cnt = 0;
  memset(vis, 0, sizeof(vis));
  vis[1] = 1;
  for (int i = 2; i < n; i++) {
    if (!vis[i]) prime[cnt++] = i;
    for (int j = 0; j < cnt && i * prime[j] < n; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  Prime(MAXN - 1);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
  }
  for (int i = 1; i <= n; i++) {
    long long tmp = sqrt(a[i]);
    if (tmp * tmp != a[i]) {
      cout << "NO" << endl;
    } else if (vis[tmp]) {
      cout << "NO" << endl;
    } else
      cout << "YES" << endl;
  }
}