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;
vector<long long int> a;
set<int> p;
int pm[int(1e6 + 1)];
void sieve() {
  for (int i = 2; i < int(1e6 + 1); i++)
    if (!pm[i]) {
      p.insert(i);
      for (int j = i; j < int(1e6 + 1); j += i) pm[j] = 1;
    }
  return;
}
void inp() {
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  return;
}
bool check(long long int x) {
  long long int m = sqrt(x);
  if (p.count(m) && m * m == x) return true;
  return false;
}
void solve() {
  sieve();
  for (auto i : a)
    if (check(i))
      cout << "YES\n";
    else
      cout << "NO\n";
  return;
}
int main() {
  clock_t beg = clock();
  {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
  };
  inp();
  solve();
  clock_t end = clock();
  fprintf(stderr, "%.3f sec\n", (double)(end - beg) / CLOCKS_PER_SEC);
  return 0;
}