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 n;
struct comp {
  bool operator()(const int &a, const int &b) { return a < b; }
};
int last = 1000001;
bool p[1000005];
set<long long> valid;
void primes() {
  memset(p, true, sizeof(p));
  p[1] = false;
  for (int i = 2; i * i < last; i++) {
    if (!p[i]) continue;
    for (int j = i * i; j < last; j += i) {
      p[j] = false;
    }
  }
  for (int i = int(1); i <= int(last); i++) {
    if (p[i]) {
      valid.insert((long long)i * i);
    }
  }
}
long long num;
int main() {
  int T;
  scanf("%d", &T);
  float f;
  primes();
  for (int i = int(1); i <= int(T); i++) {
    scanf("%I64d", &num);
    if (valid.find(num) != valid.end()) {
      cout << "YES" << endl;
    } else
      cout << "NO" << endl;
  }
}