#include <bits/stdc++.h>
using namespace std;
const int INF = INT_MAX / 3;
const double PI = acos(-1.0);
const long long MOD = 1000000007;
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
long long M;
long long F(long long N, long long P) {
  if (P == 0) return 1;
  if (P % 2 == 0) {
    long long ret = F(N, P / 2);
    return ((ret % M) * (ret % M)) % M;
  } else
    return ((N % M) * (F(N, P - 1) % M)) % M;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  if (n % 2 == 0) {
    int k = 0;
    while (n != 0) {
      n = n - 2;
      k++;
    }
    cout << k << "\n";
    for (int i = 0; i < k - 1; i++) {
      cout << 2 << " ";
    }
    cout << "2\n";
  } else {
    int k = 0;
    n = n - 3;
    while (n != 0) {
      n = n - 2;
      k++;
    }
    cout << k + 1 << "\n";
    for (int i = 0; i < k; i++) {
      cout << 2 << " ";
    }
    cout << "3\n";
  }
  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;
const int INF = INT_MAX / 3;
const double PI = acos(-1.0);
const long long MOD = 1000000007;
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
long long M;
long long F(long long N, long long P) {
  if (P == 0) return 1;
  if (P % 2 == 0) {
    long long ret = F(N, P / 2);
    return ((ret % M) * (ret % M)) % M;
  } else
    return ((N % M) * (F(N, P - 1) % M)) % M;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  if (n % 2 == 0) {
    int k = 0;
    while (n != 0) {
      n = n - 2;
      k++;
    }
    cout << k << "\n";
    for (int i = 0; i < k - 1; i++) {
      cout << 2 << " ";
    }
    cout << "2\n";
  } else {
    int k = 0;
    n = n - 3;
    while (n != 0) {
      n = n - 2;
      k++;
    }
    cout << k + 1 << "\n";
    for (int i = 0; i < k; i++) {
      cout << 2 << " ";
    }
    cout << "3\n";
  }
  return 0;
}

```