#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);
}
int Min(int a, int b, int c) {
  a = a < b ? a : b;
  a = a < c ? a : c;
  return a;
}
int main() {
  long long x, factor;
  cin >> x;
  if (x < 4) {
    cout << 1 << endl;
    cout << x << endl;
  } else if (x % 2 == 0) {
    factor = x / 2;
    cout << factor << endl;
    for (int i = 0; i < factor; i++) {
      cout << 2 << " ";
    }
    cout << endl;
  } else {
    x--;
    factor = x / 2;
    cout << factor << endl;
    for (int i = 0; i < factor - 1; i++) {
      cout << 2 << " ";
    }
    cout << 3 << endl;
  }
  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;
long long Gcd(long long a, long long b) {
  if (b == 0) return a;
  return Gcd(b, a % b);
}
int Min(int a, int b, int c) {
  a = a < b ? a : b;
  a = a < c ? a : c;
  return a;
}
int main() {
  long long x, factor;
  cin >> x;
  if (x < 4) {
    cout << 1 << endl;
    cout << x << endl;
  } else if (x % 2 == 0) {
    factor = x / 2;
    cout << factor << endl;
    for (int i = 0; i < factor; i++) {
      cout << 2 << " ";
    }
    cout << endl;
  } else {
    x--;
    factor = x / 2;
    cout << factor << endl;
    for (int i = 0; i < factor - 1; i++) {
      cout << 2 << " ";
    }
    cout << 3 << endl;
  }
  return 0;
}

```