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;
long long n, p, cntr, sq;
long long isprime[(int)1e6 + 1], lookout[(int)1e6 + 1];
vector<long long> v;
void sieve() {
  fill(isprime, isprime + (int)1e6 + 1, 1);
  isprime[0] = isprime[1] = 0;
  for (int i = 4; i < (int)1e6 + 1; i += 2) isprime[i] = 0;
  for (long long i = 3; i < (int)1e6 + 1; ++i)
    if (isprime[i])
      for (long long j = i * i; j < (int)1e6 + 1; j += 2 * i) {
        isprime[j] = 0;
      }
}
int main() {
  cin >> n;
  sieve();
  while (n--) {
    cin >> p;
    sq = sqrt(p);
    if (isprime[sq] && sq * sq == p)
      printf("YES\n");
    else
      printf("NO\n");
    cntr = 0;
  }
  return 0;
}