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: Wrong Answer)

// BBQ Easy

#include<stdio.h>
#include<stdlib.h>

void exch(int *a, int *b);
void bsort(int *l, int len);
int min(int a, int b);

int main(int argc, char *argv[]) {
  int n, tmp, sum = 0;
  int *l;

  n = (int)argv[1];
  l = malloc(sizeof(int) * 2 * n);
  for(tmp = 2; tmp <= argc; tmp++) {
    l[tmp-2] = (int)argv[tmp];
  }
  bsort(l, n);
  for(tmp = 0; tmp <= n; tmp = tmp+2) {
    sum += l[tmp];
  }
  printf("%d\n", sum);
  
  return 0;
}

void exch(int *a, int *b) {
  *a = *a + *b;
  *b = *a - *b;
  *a = *a - *b;
}

void bsort(int *l, int len) {
  int i, j;
  for(i = 0; i < len; i++) {
    for(j = i+1; j < len; l++) {
      if(l[i] < l[j]) exch(l+i, l+j);
    }
  }
}