#include<cstdio>
int n,a[1000],cnt,ans;
int main()
{
	scanf("%d",&n);
	while(n) {a[++cnt]=n%10;n/=10;ans+=a[cnt];}
	if(ans==1&&cnt>1) printf("%d\n",10);else printf("%d\n",ans);
	return 0;
}
### Prompt

Please provide a cpp coded solution 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<cstdio>
int n,a[1000],cnt,ans;
int main()
{
	scanf("%d",&n);
	while(n) {a[++cnt]=n%10;n/=10;ans+=a[cnt];}
	if(ans==1&&cnt>1) printf("%d\n",10);else printf("%d\n",ans);
	return 0;
}
```