#include<bits/stdc++.h>
using namespace std;

long long int N, a;

int main(){
    scanf("%lld", &N);
    while(N){
        a += N % 10;
        N /= 10;
    }
    if(a == 1)a = 10;
    printf("%lld", a);
    return 0;
}
### Prompt

Create a solution in cpp for 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<bits/stdc++.h>
using namespace std;

long long int N, a;

int main(){
    scanf("%lld", &N);
    while(N){
        a += N % 10;
        N /= 10;
    }
    if(a == 1)a = 10;
    printf("%lld", a);
    return 0;
}
```