#include <bits/stdc++.h>
#include <math.h>
using namespace std;
int main() {
 int N,k=0,d,s=0;
  cin>>N;
  int n=N;
  while(n){
    k++;
  	n/=10;
  }
  N-=pow(10,k-2);
  while(N){
  	d=N%10;
    s+=d;
    N/=10;
  }
  cout<<s+1<<endl;
  
}	

### Prompt

Please formulate 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 <bits/stdc++.h>
#include <math.h>
using namespace std;
int main() {
 int N,k=0,d,s=0;
  cin>>N;
  int n=N;
  while(n){
    k++;
  	n/=10;
  }
  N-=pow(10,k-2);
  while(N){
  	d=N%10;
    s+=d;
    N/=10;
  }
  cout<<s+1<<endl;
  
}	

```