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;
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1000010;
const int mod = 998244353;
long long n, k;
long long a[maxn];
set<long long> s;
int v[maxn];
void primes(long long n) {
  memset(v, 0, sizeof v);
  for (int i = 2; i <= n; ++i) {
    if (v[i]) continue;
    for (int j = i; j <= n / i; ++j) v[i * j] = 1;
  }
}
int main() {
  primes(1000000);
  for (long long i = 2; i <= 1000000; ++i) {
    if (!v[i]) s.insert(i * i);
  }
  scanf("%lld", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  for (int i = 1; i <= n; ++i) {
    if (s.find(a[i]) != s.end())
      printf("YES\n");
    else
      printf("NO\n");
  }
}