#include<stdio.h>
int main()
{
	int n;
	int sum=0;
	scanf("%d",&n);
	while(n>0)
	{
		sum+=n%10;
		n/=10; 
	 } 
	 if(sum==1)
	 {
	 	printf("10");
	 	return 0;
	 }
	 else 
	 {
	 	printf("%d",sum);
	 }
 } 
### 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 main()
{
	int n;
	int sum=0;
	scanf("%d",&n);
	while(n>0)
	{
		sum+=n%10;
		n/=10; 
	 } 
	 if(sum==1)
	 {
	 	printf("10");
	 	return 0;
	 }
	 else 
	 {
	 	printf("%d",sum);
	 }
 } 
```