#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long factorial(long long a) {
  if (a == 1) return 1;
  return a * factorial(a - 1);
}
bool ispalindrome(string s) {
  string s1 = s;
  reverse(s1.begin(), s1.end());
  if (s1.compare(s) == 0)
    return 1;
  else
    return 0;
}
void tarea() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if ((n & 1) == 0) {
    cout << n / 2 << "\n";
    while (n) {
      cout << "2"
           << " ";
      n -= 2;
    }
    cout << "\n";
  } else {
    cout << n / 2 << "\n";
    for (long long i = 1; i < (n / 2); i++) {
      cout << "2"
           << " ";
    }
    cout << "3"
         << "\n";
  }
  return;
}
int main() {
  tarea();
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long factorial(long long a) {
  if (a == 1) return 1;
  return a * factorial(a - 1);
}
bool ispalindrome(string s) {
  string s1 = s;
  reverse(s1.begin(), s1.end());
  if (s1.compare(s) == 0)
    return 1;
  else
    return 0;
}
void tarea() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if ((n & 1) == 0) {
    cout << n / 2 << "\n";
    while (n) {
      cout << "2"
           << " ";
      n -= 2;
    }
    cout << "\n";
  } else {
    cout << n / 2 << "\n";
    for (long long i = 1; i < (n / 2); i++) {
      cout << "2"
           << " ";
    }
    cout << "3"
         << "\n";
  }
  return;
}
int main() {
  tarea();
  return 0;
}

```