#include <bits/stdc++.h>
using namespace std;
template <class T>
bool umin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool umax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
bool compare(T x, U y) {
  return (abs(x - y) <= 1e-9);
}
void solve() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  if (n == 2)
    cout << 1 << "\n" << 2;
  else if (n == 3)
    cout << 1 << "\n" << 3;
  else {
    if (n & 1) {
      int k = 1;
      k += (n - 3) / 2;
      cout << k << "\n";
      cout << 3 << " ";
      k--;
      for (int i = 1; i <= k; i++) cout << 2 << " ";
    } else {
      cout << n / 2 << "\n";
      for (int i = 1; i <= (n / 2); i++) cout << 2 << " ";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
}

### 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;
template <class T>
bool umin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool umax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
bool compare(T x, U y) {
  return (abs(x - y) <= 1e-9);
}
void solve() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  if (n == 2)
    cout << 1 << "\n" << 2;
  else if (n == 3)
    cout << 1 << "\n" << 3;
  else {
    if (n & 1) {
      int k = 1;
      k += (n - 3) / 2;
      cout << k << "\n";
      cout << 3 << " ";
      k--;
      for (int i = 1; i <= k; i++) cout << 2 << " ";
    } else {
      cout << n / 2 << "\n";
      for (int i = 1; i <= (n / 2); i++) cout << 2 << " ";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
}

```