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 int S = 1e6 + 5;
int n, r, l;
vector<int> v;
bitset<S> isComp;
void seive() {
  isComp[0] = isComp[1] = 1;
  for (long long i = 2; i * i <= S; ++i)
    if (!isComp[i])
      for (int j = i * i; j <= S; j += i) isComp[j] = 1, v.push_back(j);
}
int main() {
  seive();
  sort(v.begin(), v.end());
  scanf("%d", &n);
  r = upper_bound(v.begin(), v.end(), n) - v.begin();
  while (l < r) {
    if (v[r] + v[l] == n) return printf("%d %d\n", v[l], v[r]), 0;
    (v[r] + v[l] > n) ? r-- : l++;
  }
}
