#include <bits/stdc++.h>
int main() {
  int n, k;
  int i;
  scanf("%d", &n);
  if (n % 2 == 0) {
    k = n / 2;
    printf("%d\n", k);
    for (i = 0; i < k - 1; i++) {
      printf("2 ");
    }
    printf("2");
  } else {
    k = (n - 1) / 2;
    printf("%d\n", k);
    for (i = 0; i < k - 1; i++) {
      printf("2 ");
    }
    printf("3");
  }
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.

Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k. 

Input

The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).

Output

The first line of the output contains a single integer k — maximum possible number of primes in representation.

The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.

Examples

Input

5


Output

2
2 3


Input

6


Output

3
2 2 2

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  int n, k;
  int i;
  scanf("%d", &n);
  if (n % 2 == 0) {
    k = n / 2;
    printf("%d\n", k);
    for (i = 0; i < k - 1; i++) {
      printf("2 ");
    }
    printf("2");
  } else {
    k = (n - 1) / 2;
    printf("%d\n", k);
    for (i = 0; i < k - 1; i++) {
      printf("2 ");
    }
    printf("3");
  }
  return 0;
}

```