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

### Prompt

Your challenge is to write 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 = 0, j, k = 0;
  scanf("%d", &n);
  while (1) {
    i = i + 2;
    k++;
    if (i == n) {
      printf("%d\n", k);
      break;
    }
    if (i > n) {
      printf("%d\n", k - 1);
      break;
    }
  }
  if (i == n) {
    for (j = 1; j <= k; j++) {
      printf("2 ");
    }
  } else {
    for (j = 1; j <= k - 2; j++) {
      printf("2 ");
    }
    printf("3");
  }
}

```