#include<bits/stdc++.h>
using namespace std;
int digit(int n){
int s=0;
while(n){
s+=(n%10);
n/=10;
}
return s;
}
int main(){
int n, i, j, min=INT_MAX, s;
cin>>n;
for(i=1; i<=n/2; i++){
j=n-i;
s=digit(i)+digit(j);
if(s<min){
min=s;
}
}
cout<<min;
return 0;
}
### 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;
int digit(int n){
int s=0;
while(n){
s+=(n%10);
n/=10;
}
return s;
}
int main(){
int n, i, j, min=INT_MAX, s;
cin>>n;
for(i=1; i<=n/2; i++){
j=n-i;
s=digit(i)+digit(j);
if(s<min){
min=s;
}
}
cout<<min;
return 0;
}
```