#include <stdio.h>
int main(){
    int n;
    scanf("%d",&n);
    if(n==10||n==100||n==1000||n==10000||n==100000)printf("%d\n",10);
    else printf("%d\n",n/10000+n/1000%10+n/100%10+n/10%10+n%10);
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
    scanf("%d",&n);
    if(n==10||n==100||n==1000||n==10000||n==100000)printf("%d\n",10);
    else printf("%d\n",n/10000+n/1000%10+n/100%10+n/10%10+n%10);
}
```