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;
using namespace std;
const int SIEVE_MAXN = 10000001;
class Sieve {
 private:
  int n;
  vector<long long> primes;
  bool* mark;

 public:
  Sieve(int n = SIEVE_MAXN) {
    this->n = n;
    mark = new bool[n + 1];
    fill(mark, mark + n, 1);
    for (int i = 2; i <= n; i++) {
      if (mark[i]) {
        primes.push_back(i);
        for (int j = 2; j * i <= n; j++) {
          mark[j * i] = false;
        }
      }
    }
    free(mark);
  }
  vector<long long> get_primes() { return primes; }
};
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
int XMOV[4] = {0, 0, 1, -1};
int YMOV[4] = {1, -1, 0, 0};
using namespace std;
class BTPrimes {
 public:
  void solve(std::istream& in, std::ostream& out) {
    long long x;
    int n;
    in >> n;
    Sieve s(1000001);
    vector<long long> primes = s.get_primes();
    set<long long> tprimes;
    for (int i = (0); i <= ((int)(primes.size()) - 1); i++)
      tprimes.insert(((primes[i]) * (primes[i])));
    for (int i = (0); i <= ((int)(n)-1); i++) {
      in >> x;
      if (tprimes.count(x))
        out << "YES\n";
      else
        out << "NO\n";
    }
  }
};
int main() {
  BTPrimes solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}