#include<iostream>
using namespace std;
const int maxn = 1e6+10;
int m,n;
int main()
{
	cin>>n;
	int ans=0;
	if(n%10!=0)
	{
		while(n)
		{
			ans+=n%10;
			n/=10;
		}
	}
	else 
	{
		n/=2;
		while(n)
		{
			ans+=n%10;
			n/=10;
		}
		ans*=2;
	}
	cout<<ans;
}
### 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<iostream>
using namespace std;
const int maxn = 1e6+10;
int m,n;
int main()
{
	cin>>n;
	int ans=0;
	if(n%10!=0)
	{
		while(n)
		{
			ans+=n%10;
			n/=10;
		}
	}
	else 
	{
		n/=2;
		while(n)
		{
			ans+=n%10;
			n/=10;
		}
		ans*=2;
	}
	cout<<ans;
}
```