#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, j, k = 0, s = 0, c = 0, arr[10000], var[10000], l, n;
  string x;
  cin >> n;
  if (n % 2 == 0) {
    printf("%d\n", n / 2);
    for (i = 0; i < n / 2; i++) {
      printf("%d ", 2);
    }
  } else {
    printf("%d\n", ((n - 3) / 2) + 1);
    for (i = 0; i < (n / 2) - 1; i++) {
      cout << 2 << " ";
    }
    cout << 3;
  }
}

### 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>
using namespace std;
int main() {
  int i, j, k = 0, s = 0, c = 0, arr[10000], var[10000], l, n;
  string x;
  cin >> n;
  if (n % 2 == 0) {
    printf("%d\n", n / 2);
    for (i = 0; i < n / 2; i++) {
      printf("%d ", 2);
    }
  } else {
    printf("%d\n", ((n - 3) / 2) + 1);
    for (i = 0; i < (n / 2) - 1; i++) {
      cout << 2 << " ";
    }
    cout << 3;
  }
}

```