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;
void input_output() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
bool is_prime(int n) {
  if (n < 2) return false;
  int i;
  for (i = 2; i < (int)sqrt(n) + 1; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
bool is_palindrome(string s) {
  int i;
  for (i = 0; i < s.size() / 2; i++) {
    if (s[i] != s[s.size() - i - 1]) return false;
  }
  return true;
}
int no_of_primes = 1000000;
vector<bool> prime(no_of_primes + 1, true);
vector<int> prime_nums;
void sieve_of_erat() {
  prime[0] = false;
  prime[1] = false;
  int i, j;
  for (i = 2; i < (int)sqrt(no_of_primes) + 1; i++) {
    if (prime[i] == false) continue;
    j = 2 * i;
    while (j <= no_of_primes) {
      prime[j] = false;
      j += i;
    }
  }
  for (i = 0; i < no_of_primes + 1; i++) {
    if (prime[i] == true) prime_nums.push_back(i);
  }
}
vector<vector<int> > adj(1001);
int bfs(int si, int n) {
  vector<bool> visited(n + 1, false);
  vector<int> dist(n + 1, -1);
  dist[si] = 0;
  queue<int> q;
  q.push(si);
  int i;
  while (!q.empty()) {
    int temp = q.front();
    q.pop();
    for (i = 0; i < adj[temp].size(); i++) {
      if (!visited[adj[temp][i]]) {
        q.push(adj[temp][i]);
        dist[adj[temp][i]] = dist[temp] + 1;
        visited[adj[temp][i]] = true;
      }
    }
  }
  return *max_element(dist.begin(), dist.end());
}
void solve() {
  int n;
  cin >> n;
  int num1, num2;
  int i;
  for (i = 2; i < n / 2 + 1; i++) {
    if (!prime[i] && !prime[n - i]) {
      cout << i << ' ' << n - i;
      break;
    }
  }
}
int main() {
  input_output();
  sieve_of_erat();
  solve();
}
