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 long long MAX = 1000001;
vector<bool> prime(MAX, true);
void sieve(vector<bool> &prime) {
  prime[0] = prime[1] = false;
  for (int i = 2; i * i < MAX; i++) {
    if (prime[i]) {
      for (int j = i * i; j < MAX; j += i) prime[j] = false;
    }
  }
}
long long fast_pow(long long a, long long p) {
  long long res = 1;
  while (p) {
    if (p % 2 == 0) {
      a = (a * a) % 1000000007;
      p /= 2;
    } else {
      res = (res * a) % 1000000007;
      p--;
    }
  }
  return res;
}
int expFactor(int n, int p) {
  int x = p;
  int exponent = 0;
  while ((n / x) > 0) {
    exponent += n / x;
    x *= p;
  }
  return exponent;
}
long long Sqrt(long long x) {
  if (x == 0 || x == 1) return x;
  long long i = 1, result = 1;
  while (result <= x) {
    i++;
    result = i * i;
  }
  return i - 1;
}
void c_p() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int check_perfect_square(long long n) {
  double sqrt_n = sqrt(n);
  if (sqrt_n == int(sqrt_n))
    return (int)(sqrt_n);
  else
    return -1;
}
int main() {
  c_p();
  sieve(prime);
  long long n, a;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    int t = check_perfect_square(a);
    if (t != -1 && prime[t])
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}