#include <bits/stdc++.h>
using namespace std;
int n;
const int maxn = 20000 + 5;
int cow[maxn];
int compare(int a, int b) { return a > b; }
int main(void) {
  int n;
  cin >> n;
  int num = 0;
  vector<int> arr;
  while (n != 0) {
    if (n == 3) {
      n = n - 3;
      arr.push_back(3);
    } else {
      n = n - 2;
      arr.push_back(2);
    }
    num++;
  }
  std::cout << num << std::endl;
  for (int i = 0; i < arr.size(); i++) {
    std::cout << arr[i] << std::endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 n;
const int maxn = 20000 + 5;
int cow[maxn];
int compare(int a, int b) { return a > b; }
int main(void) {
  int n;
  cin >> n;
  int num = 0;
  vector<int> arr;
  while (n != 0) {
    if (n == 3) {
      n = n - 3;
      arr.push_back(3);
    } else {
      n = n - 2;
      arr.push_back(2);
    }
    num++;
  }
  std::cout << num << std::endl;
  for (int i = 0; i < arr.size(); i++) {
    std::cout << arr[i] << std::endl;
  }
  return 0;
}

```