#include<iostream>
int f(int x){int res=0;while(x)res+=x%10,x/=10;return res;}int main(){int n,res=1e9;std::cin>>n;for(int a=1,b;a<n;a++){b=n-a;int sum=f(a)+f(b);if(res>sum)res=sum;}std::cout<<res;}

### 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<iostream>
int f(int x){int res=0;while(x)res+=x%10,x/=10;return res;}int main(){int n,res=1e9;std::cin>>n;for(int a=1,b;a<n;a++){b=n-a;int sum=f(a)+f(b);if(res>sum)res=sum;}std::cout<<res;}

```