Problem:
Score : 200 points

Problem Statement

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".

Sample Input 1

15

Sample Output 1

6

When A=2 and B=13, the sums of their digits are 2 and 4, which minimizes the value in question.

Sample Input 2

100000

Sample Output 2

10

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <math.h>
#include <limits.h>
#include <assert.h>
#define FOR(i,a,n) for(i=(a);i<(n);i++)
#define swap(type,a,b) do{type t=a;a=b;b=t;}while(0)
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))
#define ll long long
#define INF 100000000
#define MOD 1000000007
#define SIZE 100005
#define LOCAL
#ifdef LOCAL
    #define DEBUG(...) printf(" "__VA_ARGS__)
#else
    #define DEBUG(...)
#endif /* LOCAL */
int comp(const void* a,const void* b){
	return *(int*)a-*(int*)b;
}
int digitsum(int n){
	int res=0;
	while(n>0){
		res+=n%10;
		n/=10;
	}
	return res;
}
int main(void)
{
	int n,res=INF,t,i;
	scanf("%d",&n);
	FOR(i,1,n){
		t=digitsum(i)+digitsum(n-i);
		res=MIN(res,t);
	}
	printf("%d\n",res);
	return 0;
}