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.
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1000000000 + 7;
const double esp = 1e-13;
const double pi = 3.141592653589;
long long power(long long a, long long n) {
  long long tmp, c;
  tmp = c = a;
  n--;
  while (n) {
    if (n % 2) tmp = tmp * c;
    c = c * c;
    n /= 2;
  }
  return tmp;
}
int bit(long long x, int y) { return (x >> (y - 1) & 1); }
int n;
int uoc(int n) {
  int dem, n1;
  dem = 0;
  for (int j = (1), _b = ((int)(sqrt(n))); j <= _b; j++)
    if (n % j == 0) dem += 2;
  n1 = (int)sqrt(n);
  dem -= (n1 * n1 == n);
  return dem;
}
int main() {
  cin >> n;
  for (int i = (2), _b = (n); i <= _b; i++)
    if (uoc(i) > 2 && uoc(n - i) > 2) {
      cout << i << " " << n - i;
      return 0;
    }
}
