#include <bits/stdc++.h>
using namespace std;
int main() {
  long n;
  while (cin >> n) {
    long i, j, a = n, s = 0, d = 0, l = 0, f = 0;
    if (a % 2 == 0) {
      s = a / 2;
      cout << s << endl;
      while (s--) cout << "2 ";
      cout << endl;
    } else {
      s = a - 2;
      d = a - 3;
      if (s % 2 == 0) {
        l = s / 2;
        f = l + 1;
        cout << f << endl;
        while (f--) cout << "2 ";
        cout << endl;
      } else if (d % 2 == 0) {
        l = s / 2;
        f = l + 1;
        cout << f << endl;
        while (l--) cout << "2 ";
        cout << "3" << endl;
      }
    }
  }
  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;
int main() {
  long n;
  while (cin >> n) {
    long i, j, a = n, s = 0, d = 0, l = 0, f = 0;
    if (a % 2 == 0) {
      s = a / 2;
      cout << s << endl;
      while (s--) cout << "2 ";
      cout << endl;
    } else {
      s = a - 2;
      d = a - 3;
      if (s % 2 == 0) {
        l = s / 2;
        f = l + 1;
        cout << f << endl;
        while (f--) cout << "2 ";
        cout << endl;
      } else if (d % 2 == 0) {
        l = s / 2;
        f = l + 1;
        cout << f << endl;
        while (l--) cout << "2 ";
        cout << "3" << endl;
      }
    }
  }
  return 0;
}

```