#include<cstdio>
using namespace std;

int main(){
	int n;
	scanf("%d",&n);
	int t=10;
	for(int i=1;i<=5;i++){
		if(n==t){
			printf("10\n");
			return 0;
		}
		t*=10;
	}
	int ans=0;
	while(n){
		ans+=n%10;
		n/=10;
	}
	printf("%d\n",ans);
	return 0;
}
### Prompt

Generate 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>
using namespace std;

int main(){
	int n;
	scanf("%d",&n);
	int t=10;
	for(int i=1;i<=5;i++){
		if(n==t){
			printf("10\n");
			return 0;
		}
		t*=10;
	}
	int ans=0;
	while(n){
		ans+=n%10;
		n/=10;
	}
	printf("%d\n",ans);
	return 0;
}
```