#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000000;
void Search(string pat, string text) {
  int M = ((int)(pat.size())), N = ((int)(text.size()));
  for (int i = 0; i <= N - M; i++) {
    int j;
    for (j = 0; j < M; j++)
      if (text[i + j] != pat[j]) break;
    if (j == M) cout << "Pattern found at index " << i << "\n";
  }
}
bool isletter(char f) { return f >= 'a' && f <= 'z'; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k = 0;
  cin >> n;
  k = n / 2;
  cout << k << "\n";
  if (n % 2 == 0)
    for (int i = 0; i < (int)k; i++) cout << 2 << " ";
  else {
    for (int i = 0; i < (int)k - 1; i++) cout << 2 << " ";
    cout << 3;
  }
  cout << endl;
  return 0;
}

### 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>
using namespace std;
const int MOD = 1000000000;
void Search(string pat, string text) {
  int M = ((int)(pat.size())), N = ((int)(text.size()));
  for (int i = 0; i <= N - M; i++) {
    int j;
    for (j = 0; j < M; j++)
      if (text[i + j] != pat[j]) break;
    if (j == M) cout << "Pattern found at index " << i << "\n";
  }
}
bool isletter(char f) { return f >= 'a' && f <= 'z'; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k = 0;
  cin >> n;
  k = n / 2;
  cout << k << "\n";
  if (n % 2 == 0)
    for (int i = 0; i < (int)k; i++) cout << 2 << " ";
  else {
    for (int i = 0; i < (int)k - 1; i++) cout << 2 << " ";
    cout << 3;
  }
  cout << endl;
  return 0;
}

```