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 MAX = 1000006;
bool isPrime[MAX];
void sieve() {
  isPrime[0] = isPrime[1] = 0;
  for (int i = 2; i * i <= MAX; i++) {
    if (isPrime[i]) {
      for (int j = i * 2; j <= MAX; j += i) isPrime[j] = 0;
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i <= MAX; ++i) isPrime[i] = 1;
  sieve();
  for (int i = 2;; ++i) {
    if (!isPrime[i] && !isPrime[n - i]) {
      printf("%d %d\n", i, n - i);
      break;
    }
  }
}