#include<bits/stdc++.h>
using namespace std;
int n,r;
int d(int x){
	int r=0;
	while(x){
		r += x%10;
		x/=10;
	}
	return r;
}
int main(){
	scanf("%d",&n);
	r = 9999;
	for(int i=1; i+i<=n; i++)
		r = min(r, d(i)+d(n-i));
	printf("%d\n",r);
	return 0;
}

### Prompt

Your task is to create 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<bits/stdc++.h>
using namespace std;
int n,r;
int d(int x){
	int r=0;
	while(x){
		r += x%10;
		x/=10;
	}
	return r;
}
int main(){
	scanf("%d",&n);
	r = 9999;
	for(int i=1; i+i<=n; i++)
		r = min(r, d(i)+d(n-i));
	printf("%d\n",r);
	return 0;
}

```