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;
template <class T>
inline T power(T b, T p) {
  return p == 0 ? 1 : b * power(b, p - 1);
}
template <class T>
inline string tostring(T n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
template <class T>
inline T toint(T n) {
  T ret;
  stringstream s(n);
  s >> ret;
  return ret;
}
struct SM {
  int u, v, sum;
  SM(int u = 0, int v = 0, int sum = 0) : u(u), v(v), sum(sum) {}
} node[100000 + 10];
vector<char> a;
int len, ans = 0;
char ch[10];
int rec(char v, int cnt) {
  if (cnt == len) {
    for (int i = 0; i < a.size(); i++) {
      int c = 0;
      if (a[i] == ch[0]) {
        for (int j = i; j < a.size(); j++) {
          if (a[i] == a[j])
            c += 1;
          else
            break;
        }
      }
      if (c > 1) {
        ans += 1;
        break;
      }
    }
    return 0;
  }
  for (char c = 'a'; c <= 'z'; c++) {
    a.push_back(c);
    rec(c, cnt + 1);
    a.pop_back();
  }
  return 0;
}
int n;
bool F(int a) {
  for (int i = 2; i <= sqrt(a); i++) {
    if (a % i == 0) return true;
  }
  return false;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 4; i < n; i++) {
      int a = F(n - i);
      int b = F(i);
      if (a == true && b == true) {
        printf("%d %d\n", i, (n - i));
        break;
      }
    }
  }
  return 0;
}