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

### Prompt

Your task is to create 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, count;
  while (~scanf("%d", &n)) {
    i = 0, count = 0;
    while (i <= n) {
      i += 2;
      count++;
    }
    i -= 2;
    count--;
    if (i == n) {
      printf("%d\n", count);
      while (count--) {
        printf("2 ");
      }
      putchar('\n');
    } else {
      if (n - i == 1) {
        i -= 2;
      }
      printf("%d\n", count);
      count--;
      while (count--) {
        printf("2 ");
      }
      printf("%d\n", 3);
    }
  }
  return 0;
}

```