#include<iostream>
int main(){
int n,sum;
  std::cin>>n;
  for(int i=1;i<=n;i*=10){
    sum+=(n%(10*i))/i;
  }
  if(sum==1&&n!=1){
  std::cout<<10<<std::endl;
  }
  else{
  std::cout<<sum<<std::endl;
  }
}

### Prompt

Please create a solution in cpp to the following problem:

Takahashi has two positive integers A and B.

It is known that A plus B equals N. Find the minimum possible value of "the sum of the digits of A" plus "the sum of the digits of B" (in base 10).

Constraints

* 2 ≤ N ≤ 10^5
* N is an integer.

Input

Input is given from Standard Input in the following format:


N


Output

Print the minimum possible value of "the sum of the digits of A" plus "the sum of the digits of B".

Examples

Input

15


Output

6


Input

100000


Output

10

### Response

```cpp
#include<iostream>
int main(){
int n,sum;
  std::cin>>n;
  for(int i=1;i<=n;i*=10){
    sum+=(n%(10*i))/i;
  }
  if(sum==1&&n!=1){
  std::cout<<10<<std::endl;
  }
  else{
  std::cout<<sum<<std::endl;
  }
}

```