#include <bits/stdc++.h>
using namespace std;
const long long MAX = 2e5 + 10;
long long A[MAX];
void solve() {
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2 << endl;
    for (long long i = 1; i <= n / 2; ++i) cout << "2 ";
    cout << endl;
  } else {
    if (n == 3)
      cout << "1" << endl << "3" << endl;
    else {
      cout << (n / 2) << endl;
      n -= 3;
      for (long long i = 1; i <= n / 2; ++i) cout << "2 ";
      cout << "3" << endl;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T;
  solve();
  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 long long MAX = 2e5 + 10;
long long A[MAX];
void solve() {
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2 << endl;
    for (long long i = 1; i <= n / 2; ++i) cout << "2 ";
    cout << endl;
  } else {
    if (n == 3)
      cout << "1" << endl << "3" << endl;
    else {
      cout << (n / 2) << endl;
      n -= 3;
      for (long long i = 1; i <= n / 2; ++i) cout << "2 ";
      cout << "3" << endl;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T;
  solve();
  return 0;
}

```