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;
int main() {
  int a;
  cin >> a;
  int b = 0, c = 0;
  for (int i = 1; i <= a / 2; i++) {
    for (int j = 1; j <= sqrt(i); j++) {
      if (i % j == 0 && j * j != i) {
        b += 2;
      }
      if (j * j == i) {
        b++;
      }
      if (b > 3) {
        break;
      }
    }
    for (int k = 1; k <= sqrt(a - i); k++) {
      if ((a - i) % k == 0 && k * k != (a - i)) {
        c += 2;
      }
      if (k * k == (a - i)) {
        c++;
      }
      if (c > 3) {
        break;
      }
    }
    if (b >= 3 && c >= 3) {
      cout << i << ' ' << a - i;
      return 0;
    }
    b = 0;
    c = 0;
  }
}