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>
bool seive[1000009];
void is_prime();
long long a, b, c, d, i, j;
int main() {
  is_prime();
  long long n, num;
  scanf("%I64d", &num);
  n = 0;
  for (i = 2; i < (num); i++) {
    if (seive[i] == true && seive[num - i] == true) {
      n = i;
      break;
    }
  }
  if (n != 0) printf("%I64d %I64d\n", n, num - n);
}
void is_prime() {
  int k = 0, g;
  for (i = 2; i < sqrt(1000009); i++) {
    if (i == 2 || seive[i] == false) {
      for (j = i + i; j < 1000009; j = j + i) {
        seive[j] = true;
      }
    }
  }
}
