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;
vector<int> a;
int a1[1000001];
int main() {
  long long i, j, n, x = 4, y, ck = 0, ckk, root = sqrt(1000001);
  a.push_back(2);
  for (i = 4; i <= 1000000; i += 2) a1[i] = 1;
  for (i = 3; i <= 1000000; i++) {
    if (a1[i] == 0) {
      a.push_back(i);
      if (i <= root)
        for (j = i * i; j <= 1000000; j += 2 * i) a1[j] = 1;
    }
  }
  cin >> n;
  while (ck == 0) {
    y = n - x;
    for (i = 0; i <= a.size(); i++) {
      if (a[i] != x && a[i] != y)
        ckk = 1;
      else {
        ckk = 0;
        x++;
        break;
      }
    }
    if (ckk == 1) {
      cout << x << " " << y << endl;
      ck = 1;
    }
  }
  return 0;
}