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

### Prompt

Please provide a Cpp coded solution 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 ifeven(int n) {
  if (n % 2 == 0)
    return 1;
  else
    return 0;
}
int main() {
  int i, n, k, l, r = 0;
  scanf("%d", &n);
  if (ifeven(n)) {
    while (n != 0) {
      r++;
      n -= 2;
    }
    printf("%d\n", r);
    for (i = 0; i < r; i++) {
      printf("2 ");
    }
  } else {
    r++;
    n -= 3;
    while (n != 0) {
      r++;
      n -= 2;
    }
    printf("%d\n", r);
    for (i = 1; i < r; i++) {
      printf("2 ");
    }
    printf("3 ", r);
  }
  return 0;
}

```