Problem:
Score : 200 points

Problem Statement

Snuke is having a barbeque party.

At the party, he will make N servings of Skewer Meal.

Example of a serving of Skewer Meal

He has a stock of 2N skewers, all of which will be used in Skewer Meal. The length of the i-th skewer is L_i.
Also, he has an infinite supply of ingredients.

To make a serving of Skewer Meal, he picks 2 skewers and threads ingredients onto those skewers.
Let the length of the shorter skewer be x, then the serving can hold the maximum of x ingredients.

What is the maximum total number of ingredients that his N servings of Skewer Meal can hold, if he uses the skewers optimally?

Constraints

1≦N≦100

1≦L_i≦100

For each i, L_i is an integer.

Input

The input is given from Standard Input in the following format:

N
L_1 L_2 ... L_{2N}

Output

Print the maximum total number of ingredients that Snuke's N servings of Skewer Meal can hold.

Sample Input 1

2
1 3 1 2

Sample Output 1

3

If he makes a serving using the first and third skewers, and another using the second and fourth skewers, each serving will hold 1 and 2 ingredients, for the total of 3.

Sample Input 2

5
100 1 2 3 14 15 58 58 58 29

Sample Output 2

135

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

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void){
        int n,i,j,tmp;
        scanf("%d", &n);

        int l[2*n];
        char str[1024];
        char *adr;

        while( getchar() != '\n');
        if(fgets(str, 1024, stdin) == NULL){
                printf("No String");
        }

        adr = strtok(str, " ");
        l[0] = atoi(adr);
        for(i = 1; i < 2*n; i++){
                adr = strtok(NULL , " \n");
                if(adr != NULL){
                        l[i] = atoi(adr);
                }
        }

        for(i = 0; i < 2*n; i++){
                for(j = i + 1; j < 2*n; j++){
                        if(l[i] > l[j]){
                                tmp = l[i];
                                l[i] = l[j];
                                l[j] = tmp;
                        }
                }
        }

        tmp = 0;
        for(i = 0; i < 2*n; i+=2){
                tmp += l[i];
        }
        printf("%d\n", tmp);
        return 0;
}