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 <stdlib.h>

#define N 200
#define INT(X) (*((const int*)(X)))

static int d[N];
static int n;

static int compare(const void *a, const void *b) {
  if (INT(a) < INT(b)) return -1;
  if (INT(b) < INT(a)) return 1;
  return 0;
}

int main(void) {
  int res;
  int i;
  
  scanf("%d", &n);
  for (i = 0; i < 2 * n; ++i)
    scanf("%d", &d[i]);
  qsort(d, 2 * n, sizeof(int), compare);
  
  res = 0;
  for (i = 0; i < 2 * n; i += 2) {
    res += d[i];
  }
  printf("%d\n", res);
  
  return (0);
}