#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,ans=0;
    cin>>n;
    while(n){
      ans+=n%10;
      n/=10;
    }
  if(ans==1){cout<<10;}
  else{cout<<ans;}
}
### Prompt

Your challenge is to write 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>
using namespace std;
typedef long long ll;
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,ans=0;
    cin>>n;
    while(n){
      ans+=n%10;
      n/=10;
    }
  if(ans==1){cout<<10;}
  else{cout<<ans;}
}
```