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

```