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>
const int N = 5000005;
int prime[N] = {0}, num_prime = 0;
int isNotPrime[N] = {1, 1};
void init() {
  for (int i = 2; i < N; i++) {
    if (!isNotPrime[i]) {
      prime[num_prime++] = i;
      isNotPrime[i] = 0;
    }
    for (int j = 0; j < num_prime && i * prime[j] < N; j++) {
      isNotPrime[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  init();
  int num, a, b;
  while (~scanf("%d", &num)) {
    for (a = 4; a < num / 2 + 1; a++) {
      b = num - a;
      if (isNotPrime[a] && isNotPrime[b]) {
        printf("%d %d\n", a, b);
        break;
      }
    }
  }
}