Problem: One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.

For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.

You are given an integer n no less than 12, express it as a sum of two composite numbers.

Input

The only line contains an integer n (12 ≤ n ≤ 106).

Output

Output two composite integers x and y (1 < x, y < n) such that x + y = n. If there are multiple solutions, you can output any of them.

Examples

Input

12


Output

4 8


Input

15


Output

6 9


Input

23


Output

8 15


Input

1000000


Output

500000 500000

Note

In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.

In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
set<ll> factors(ll n) {
  set<ll> res;
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      res.insert(i);
      res.insert(n / i);
    }
  }
  return res;
}
bool is_prime(int n) {
  if (n < 2) return false;
  if (n == 2) return true;
  if (n % 2 == 0) return false;
  for (int i = 3; i <= sqrt(n); i += 2)
    if (n % i == 0) return false;
  return true;
}
int main() {
  ll n;
  cin >> n;
  ll lo = 0, mid = 0, hi = n;
  while (lo <= hi) {
    mid = lo + (hi - lo) / 2;
    if (lo + hi == n and !is_prime(lo) and !is_prime(hi) and lo != 0 and
        lo != 1) {
      cout << lo << " " << hi << endl;
      break;
    } else {
      lo += (mid / 4);
      hi -= (mid / 4);
    }
  }
  return 0;
}