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 long long mod = 1e9 + 7;
void flush() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin.sync_with_stdio(0);
}
int n, k, dp[100009];
int a[100009];
string s;
int solve(int i, bool r) {
  if (i >= n && r) return 1;
  if (i >= n) return 0;
  int p = 0, p1 = 0;
  if (s[i] == '0' && s[i + 1] == '1' || s[i] == '1' && s[i + 1] == '0')
    p = solve(i + 1, r), p1 = solve(i + 2, !r);
  else
    p = solve(i + 1, r);
  return p || p1;
}
int main() {
  flush();
  int n;
  cin >> n;
  cout << (n % 2 == 0 ? 4 : 9) << " ";
  cout << (n % 2 == 0 ? n - 4 : n - 9);
}