#include<stdio.h>
int f(int a)
{
	int s=0;
	while(a!=0)
	{
		s+=a%10;
		a/=10;	
	}
	return s;
}
int main()
{
	int n,min=100001,i;
	scanf("%d",&n);
	for(i=1;i<=n/2;i++)
	{
		if(f(i)+f(n-i)<min)
			min=f(i)+f(n-i);	
	}
	printf("%d",min);
	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<stdio.h>
int f(int a)
{
	int s=0;
	while(a!=0)
	{
		s+=a%10;
		a/=10;	
	}
	return s;
}
int main()
{
	int n,min=100001,i;
	scanf("%d",&n);
	for(i=1;i<=n/2;i++)
	{
		if(f(i)+f(n-i)<min)
			min=f(i)+f(n-i);	
	}
	printf("%d",min);
	return 0;
}
```