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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>


#define SIZE 10000


int main(void){

	int i,j,n,a,b,ad,bd,amariA,amariB,sum[SIZE],temp;
	scanf("%d",&n);
	//ad = 0;
	//bd = 0;	
	
	for(i=1;i<n;i++){
		 
		b = i;
		a = n - b;
		ad = 0;
		bd = 0;
	
		for(j=1;j<=5;j++){
			amariA = a%10;//(int)pow(10,j);
			printf("%d, ",amariA);
			amariB = b%10;//(int)pow(10,j);
			ad += amariA;
			//printf("%d\n",ad);
			bd += amariB;
			a -= amariA;//pow(10,j-1);
			a = a/10;
			printf("%d\n",a);
			b -= amariB;//pow(10,j-1);
			b = b/10;   
		}

		sum[i] = ad + bd;
		//printf("%d\n",sum[i]);
	}

	temp = 99999;
	for(i=1;i<n;i++){

		if(temp>=sum[i]){
			 temp = sum[i];
		//	printf("%d\n",temp);
		
		}
	}
	
	printf("%d",temp);

	return 0;
	
}