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 NN = 1000000 + 10;
bool is_p[NN];
vector<int> pp;
void make() {
  for (int i = 2; i < NN; i++) is_p[i] = 1;
  is_p[1] = 1;
  for (int i = 2; i < NN; i++) {
    if (is_p[i]) {
      for (long long j = (long long)(i)*i; j < NN; j += i) is_p[j] = 0;
    }
  }
  for (int i = 2; i < NN; i++)
    if (!is_p[i]) pp.push_back(i);
}
int main() {
  make();
  int n;
  while (scanf("%d", &n) == 1) {
    for (int i = 2; i * 2 <= n; i++) {
      if (!is_p[i] && !is_p[n - i]) {
        printf("%d %d\n", i, n - i);
        break;
      }
    }
  }
  return 0;
}