#include <bits/stdc++.h>
using namespace std;
long long int binpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a);
    a = a * a;
    b >>= 1;
  }
  return res;
}
bool is_prime(long long int n) {
  if (n == 1) {
    return false;
  }
  int i = 2;
  while (i * i <= n) {
    if (n % i == 0) {
      return false;
    }
    i += 1;
  }
  return true;
}
void solve() {
  int n, ok = 0;
  cin >> n;
  int count = 0;
  if (n % 2) {
    count++;
    n -= 3;
    ok = 1;
  }
  count += n / 2;
  cout << count << "\n";
  if (ok) cout << 3 << " ";
  while (n > 0) {
    cout << 2 << " ";
    n -= 2;
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long int binpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a);
    a = a * a;
    b >>= 1;
  }
  return res;
}
bool is_prime(long long int n) {
  if (n == 1) {
    return false;
  }
  int i = 2;
  while (i * i <= n) {
    if (n % i == 0) {
      return false;
    }
    i += 1;
  }
  return true;
}
void solve() {
  int n, ok = 0;
  cin >> n;
  int count = 0;
  if (n % 2) {
    count++;
    n -= 3;
    ok = 1;
  }
  count += n / 2;
  cout << count << "\n";
  if (ok) cout << 3 << " ";
  while (n > 0) {
    cout << 2 << " ";
    n -= 2;
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```