#include <bits/stdc++.h>
using namespace std;
const long long N = 100005, INF = 1e18;
void solve() {
  long long n, ans = 0, max = 0;
  cin >> n;
  ans = n / 2;
  cout << ans << '\n';
  for (long long i = (long long)0; i < ans - 1; i++) {
    cout << "2 ";
  }
  if (n % 2 == 0)
    cout << 2 << '\n';
  else
    cout << 3 << '\n';
}
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) solve();
}

### 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;
const long long N = 100005, INF = 1e18;
void solve() {
  long long n, ans = 0, max = 0;
  cin >> n;
  ans = n / 2;
  cout << ans << '\n';
  for (long long i = (long long)0; i < ans - 1; i++) {
    cout << "2 ";
  }
  if (n % 2 == 0)
    cout << 2 << '\n';
  else
    cout << 3 << '\n';
}
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) solve();
}

```