#include <bits/stdc++.h>
using namespace std;
long long int prime[2000000];
void siv() {
  long long int i, j;
  for (i = 0; i < 2000000; i++) {
    prime[i] = 1;
  }
  prime[0] = 0;
  prime[1] = 0;
  for (i = 2; i * i <= 2000000; i++) {
    if (prime[i] == 1) {
      for (j = i + i; j < 2000000; j += i) prime[j] = 0;
    }
  }
}
int main() {
  long long int n, a, b;
  siv();
  while (cin >> n) {
    siv();
    for (long long int i = 4; i < n; i += 2) {
      a = i;
      b = n - i;
      if (prime[b] == 0) {
        cout << a << " " << b << endl;
        break;
      }
    }
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int prime[2000000];
void siv() {
  long long int i, j;
  for (i = 0; i < 2000000; i++) {
    prime[i] = 1;
  }
  prime[0] = 0;
  prime[1] = 0;
  for (i = 2; i * i <= 2000000; i++) {
    if (prime[i] == 1) {
      for (j = i + i; j < 2000000; j += i) prime[j] = 0;
    }
  }
}
int main() {
  long long int n, a, b;
  siv();
  while (cin >> n) {
    siv();
    for (long long int i = 4; i < n; i += 2) {
      a = i;
      b = n - i;
      if (prime[b] == 0) {
        cout << a << " " << b << endl;
        break;
      }
    }
  }
}

```