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 read() {
  int x;
  scanf("%d", &x);
  return x;
}
int readln() {
  int x;
  scanf("%d\n", &x);
  return x;
}
const int MAX_N = int(1e6) + 6;
bool ok[MAX_N];
int main() {
  memset(ok, (true), sizeof(ok));
  ok[1] = false;
  for (long long i = 2; i < MAX_N; ++i)
    if (ok[i])
      for (long long j = i * i; j < MAX_N; j += i) ok[j] = false;
  int n = read();
  for (int i = 4; i < n; ++i)
    if (!ok[i]) {
      int j = n - i;
      if (j > 1 && !ok[j]) {
        printf("%d %d\n", i, j);
        break;
      }
    }
  return 0;
}