#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using db = double;
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, -1, 0, 1};
bool isPunct(char c) {
  string punct = ".,!?";
  for (int i = (0); i < (((int)(punct).size())); ++i) {
    if (c == punct[i]) {
      return true;
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  int n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2 << endl;
  } else {
    n -= 3;
    cout << 1 + n / 2 << endl;
    cout << 3 << " ";
  }
  for (int k = (0); k < (n / 2); ++k) {
    cout << 2 << " ";
  }
  cout << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using db = double;
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, -1, 0, 1};
bool isPunct(char c) {
  string punct = ".,!?";
  for (int i = (0); i < (((int)(punct).size())); ++i) {
    if (c == punct[i]) {
      return true;
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  int n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2 << endl;
  } else {
    n -= 3;
    cout << 1 + n / 2 << endl;
    cout << 3 << " ";
  }
  for (int k = (0); k < (n / 2); ++k) {
    cout << 2 << " ";
  }
  cout << endl;
  return 0;
}

```