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.
#include <bits/stdc++.h>
using namespace std;
void isPrime(int primes[], int l, int m, int y) {
  int i;
  for (i = 0; i < y; i++) {
    if (primes[i] == l || primes[i] == m) {
      l--;
      m++;
      return isPrime(primes, l--, m++, y);
    }
  }
  if (i == y) cout << l << " " << m;
}
int main() {
  int a[1000000] = {0}, primes[1000000 / 2];
  int y = 0, n = 0;
  for (int i = 2; i < sqrt(1000000); i++) {
    if (a[i] == 0) {
      for (int j = 2; (j * i) < 1000000; j++) {
        a[j * i] = 1;
      }
    }
  }
  for (int i = 0; i < 1000000; i++) {
    if (a[i] == 0) {
      primes[y++] = i;
    }
  }
  long long int k, l, m;
  cin >> k;
  l = k / 2;
  m = k - l;
  isPrime(primes, l, m, y);
  return 0;
}
