#include<cstdio>
using namespace std;
int main(){
	int n,ans=1e7;
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		int a=i,b=n-i,num=0;
		while (a) num+=a%10,a/=10;
		while (b) num+=b%10,b/=10;
		if (num<ans) ans=num;
	}
	printf("%d",ans);
	return 0;	
}
### Prompt

Develop a solution in cpp to the problem described below:

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<cstdio>
using namespace std;
int main(){
	int n,ans=1e7;
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		int a=i,b=n-i,num=0;
		while (a) num+=a%10,a/=10;
		while (b) num+=b%10,b/=10;
		if (num<ans) ans=num;
	}
	printf("%d",ans);
	return 0;	
}
```