#include<bits/stdc++.h>
#define REP(i,n) for(int i=0;n!=0;i++)
using namespace std;


int main(){
  int N,sum=0;cin>>N;int S=N;
  REP(i,N)sum+=N%10,N/=10;
  if(sum==1&&S>=10)sum=10;
  cout<<sum<<endl;
}
### Prompt

Develop a solution in Cpp 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<bits/stdc++.h>
#define REP(i,n) for(int i=0;n!=0;i++)
using namespace std;


int main(){
  int N,sum=0;cin>>N;int S=N;
  REP(i,N)sum+=N%10,N/=10;
  if(sum==1&&S>=10)sum=10;
  cout<<sum<<endl;
}
```