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>
const double PI = acos(-1.0);
const double e = exp(1.0);
const int INF = 0x7fffffff;
;
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
inline T Min(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline T Max(T a, T b) {
  return a > b ? a : b;
}
bool cmpbig(int a, int b) { return a > b; }
bool cmpsmall(int a, int b) { return a < b; }
using namespace std;
int heshu[1000010];
int heset[1000010];
int k;
void init() {
  memset(heshu, 0, sizeof(heshu));
  for (int i = 2; i * i <= 1e6; i++)
    for (int j = i; j * i <= 1e6; j++) heshu[i * j] = 1;
  k = 0;
  for (int i = 4; i < 1e6; i++)
    if (heshu[i]) heset[k++] = i;
}
bool check(int num) {
  int l = 0, r = k - 1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (num > heset[mid])
      l = mid + 1;
    else if (num < heset[mid])
      r = mid - 1;
    else
      return true;
  }
  return false;
}
int main() {
  int n;
  init();
  while (cin >> n) {
    for (int i = 4; i < n; i++) {
      if (heshu[i] && check(n - i)) {
        cout << i << " " << n - i << endl;
        break;
      }
    }
  }
  return 0;
}
