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;
const int maxn = 1e6 + 5;
bool prime[maxn];
void primes(int n) {
  for (int i = 2; i <= n; i++) {
    if (!prime[i]) continue;
    for (int j = i; j <= n / i; j++) prime[i * j] = false;
  }
}
int main() {
  for (int i = 1; i <= maxn; i++) prime[i] = true;
  int n;
  scanf("%d", &n);
  if (n % 4 == 0) {
    printf("%d %d\n", n / 2, n / 2);
    return 0;
  }
  primes(n);
  prime[1] = false;
  for (int i = 4; i <= n / 2; i++) {
    int j = n - i;
    if (!prime[i] && !prime[j]) {
      printf("%d %d\n", i, j);
      return 0;
    }
  }
  return 0;
}