#include<stdio.h>
int main(){
  int N,sum=0;
  scanf("%d",&N );
  while(N>0){
    sum+=N%10;
    N=N/10;
  }
  if(sum==1)
    printf("10\n");
  else
    printf("%d\n",sum );
  return 0;
}

### Prompt

Generate a Cpp solution 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<stdio.h>
int main(){
  int N,sum=0;
  scanf("%d",&N );
  while(N>0){
    sum+=N%10;
    N=N/10;
  }
  if(sum==1)
    printf("10\n");
  else
    printf("%d\n",sum );
  return 0;
}

```