#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n;
long long a[100010];
bool mark[1000010];
int main() {
  while (cin >> n) {
    memset(mark, 1, sizeof(mark));
    mark[1] = 0;
    for (int i = 2; i <= n; i++)
      if (mark[i])
        for (int j = 2 * i; j <= n; j += i) mark[j] = 0;
    for (int i = 2; i <= n; i++)
      if (!mark[i] && !mark[n - i] && i < n - i) {
        printf("%d %d\n", i, n - i);
        break;
      }
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n;
long long a[100010];
bool mark[1000010];
int main() {
  while (cin >> n) {
    memset(mark, 1, sizeof(mark));
    mark[1] = 0;
    for (int i = 2; i <= n; i++)
      if (mark[i])
        for (int j = 2 * i; j <= n; j += i) mark[j] = 0;
    for (int i = 2; i <= n; i++)
      if (!mark[i] && !mark[n - i] && i < n - i) {
        printf("%d %d\n", i, n - i);
        break;
      }
  }
  return 0;
}

```