#include <cstdio>

int main(){
    int N;
    scanf("%d", &N);
    int sum = 0;
    while (N>0){
        sum += N%10;
        N /= 10;
    }
    if (sum==1) sum = 10;
    printf("%d", sum);
    return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <cstdio>

int main(){
    int N;
    scanf("%d", &N);
    int sum = 0;
    while (N>0){
        sum += N%10;
        N /= 10;
    }
    if (sum==1) sum = 10;
    printf("%d", sum);
    return 0;
}

```