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 maxfn(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long minfn(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
void vec_cin(vector<long long> &v, long long n) {
  for (long long i = 0; i < n; i++) {
    long long buff;
    cin >> buff;
    v.push_back(buff);
  }
}
string binary_rep(long long n) {
  string s = "";
  for (long long i = 63; i >= 0; i--) {
    if (n & (1 << i))
      s += "1";
    else
      s += "0";
  }
  return s;
}
void print_vec(vector<long long> v) {
  for (auto x : v) cout << x << " ";
  cout << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n = 1e6;
  vector<char> is_prime(n + 1, true);
  is_prime[0] = is_prime[1] = false;
  for (long long i = 2; i <= n; i++) {
    if (is_prime[i] && (long long)i * i <= n) {
      for (long long j = i * i; j <= n; j += i) is_prime[j] = false;
    }
  }
  cin >> n;
  while (n--) {
    long long x;
    cin >> x;
    if (floor((double)sqrt(x)) == ceil((double)sqrt(x))) {
      if (is_prime[sqrt(x)])
        cout << "YES"
             << "\n";
      else
        cout << "NO"
             << "\n";
    } else
      cout << "NO"
           << "\n";
  }
  return 0;
}