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

### Prompt

Develop a solution in Cpp to the problem described below:

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 num, counter = 0;
  scanf("%d", &num);
  if (num % 2 == 0) {
    counter += num / 2;
    printf("%d\n", counter);
    for (int a = 0; a < counter; a++) {
      printf("2");
      if (a < counter - 1)
        printf(" ");
      else
        printf("\n");
    }
  } else if (num % 2 != 0) {
    num -= 3;
    counter++;
    counter += num / 2;
    printf("%d\n", counter);
    for (int a = 0; a < counter - 1; a++) {
      printf("2 ");
    }
    printf("3\n");
  }
  return 0;
}

```