#include <bits/stdc++.h>
using namespace std;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
int cas = 1;
int dxx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dyy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int dx[] = {-1, 0, 0, 1};
int dy[] = {0, -1, 1, 0};
int main() {
  int n;
  scanf("%d", &n);
  if (n % 2 == 0) {
    cout << n / 2 << endl;
    for (int i = 1; i <= n / 2; i++) {
      printf("2 ");
    }
    cout << endl;
  } else {
    cout << (n - 3) / 2 + 1 << endl;
    for (int i = 1; i <= (n - 3) / 2; i++) {
      printf("2 ");
    }
    printf(" 3");
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
int cas = 1;
int dxx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dyy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int dx[] = {-1, 0, 0, 1};
int dy[] = {0, -1, 1, 0};
int main() {
  int n;
  scanf("%d", &n);
  if (n % 2 == 0) {
    cout << n / 2 << endl;
    for (int i = 1; i <= n / 2; i++) {
      printf("2 ");
    }
    cout << endl;
  } else {
    cout << (n - 3) / 2 + 1 << endl;
    for (int i = 1; i <= (n - 3) / 2; i++) {
      printf("2 ");
    }
    printf(" 3");
  }
  return 0;
}

```