#include <bits/stdc++.h>
using namespace std;
bool isPrime(int num) {
  if (num == 2)
    return 1;
  else if (num == 1)
    return 0;
  else
    for (int i = 2; i <= sqrt(num); i++)
      if (num % i == 0) return 0;
  return 1;
}
int main() {
  int n, prev, k = 0;
  int a[100000];
  cin >> n;
  for (int i = 2, j = 0; i <= 3; i++) {
    if (isPrime(i)) {
      if (n % i == 0) {
        k = +n / i;
        for (int l = 0; l < (n / i); l++) {
          a[j] = i;
          j++;
        }
        break;
      } else {
        k = +n / i;
        for (int l = 0; l < (n / i); l++) {
          a[j] = i;
          j++;
        }
        a[j - 1] += 1;
        break;
      }
    }
  }
  cout << k << endl;
  for (int i = 0; i < k; i++) cout << a[i] << " ";
  return 0;
}

### 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;
bool isPrime(int num) {
  if (num == 2)
    return 1;
  else if (num == 1)
    return 0;
  else
    for (int i = 2; i <= sqrt(num); i++)
      if (num % i == 0) return 0;
  return 1;
}
int main() {
  int n, prev, k = 0;
  int a[100000];
  cin >> n;
  for (int i = 2, j = 0; i <= 3; i++) {
    if (isPrime(i)) {
      if (n % i == 0) {
        k = +n / i;
        for (int l = 0; l < (n / i); l++) {
          a[j] = i;
          j++;
        }
        break;
      } else {
        k = +n / i;
        for (int l = 0; l < (n / i); l++) {
          a[j] = i;
          j++;
        }
        a[j - 1] += 1;
        break;
      }
    }
  }
  cout << k << endl;
  for (int i = 0; i < k; i++) cout << a[i] << " ";
  return 0;
}

```