#include<cstdio>


int main(void)
{
	int n;

	while(scanf("%d", &n)==1)
	{
		int ans=0;
		for(;n>0;n/=10)
		{
			ans+=n%10;
		}
		if(ans==1) ans=10;

		printf("%d\n", ans);
	}

	return 0;
}

### Prompt

Please formulate 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<cstdio>


int main(void)
{
	int n;

	while(scanf("%d", &n)==1)
	{
		int ans=0;
		for(;n>0;n/=10)
		{
			ans+=n%10;
		}
		if(ans==1) ans=10;

		printf("%d\n", ans);
	}

	return 0;
}

```