#include<bits/stdc++.h>
using namespace std;
int d(int n){
	int ans = 0;
  	do{
    	ans+=n%10;
    }while(n/=10);
  	return ans;
}
int main(){
	int n;cin>>n;
  	int mn = 100000000;
  	for(int i=1;i<n;i++){
    	mn = min(mn,d(i)+d(n-i));
    }
  	cout<<mn;
}
### Prompt

In cpp, your task is to solve 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 d(int n){
	int ans = 0;
  	do{
    	ans+=n%10;
    }while(n/=10);
  	return ans;
}
int main(){
	int n;cin>>n;
  	int mn = 100000000;
  	for(int i=1;i<n;i++){
    	mn = min(mn,d(i)+d(n-i));
    }
  	cout<<mn;
}
```