#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    printf("%d\n", n / 2);
    if (n == 3)
      printf("3");
    else
      printf("2");
    for (int i = 2; i < n / 2; i++) printf(" 2");
    if (n % 2 == 0 && n != 2)
      printf(" 2");
    else if (n != 2 && n != 3)
      printf(" 3");
    printf("\n");
  }
}

### 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 mod = 1e9 + 7;
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    printf("%d\n", n / 2);
    if (n == 3)
      printf("3");
    else
      printf("2");
    for (int i = 2; i < n / 2; i++) printf(" 2");
    if (n % 2 == 0 && n != 2)
      printf(" 2");
    else if (n != 2 && n != 3)
      printf(" 3");
    printf("\n");
  }
}

```