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 (Clang 3.8.0) (judge verdict: Accepted)

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

int N;
int L[400];

int comp(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}
#define min(a,b) ((a > b) ? b : a)
int main(void)
{
    int i;
    int S = 0;
    scanf("%d",&N);
    for (i = 0; i < 2*N; i++)
        scanf("%d",&L[i]);
    qsort(L, 2*N, sizeof(int), comp);
    for (i = 0; i < N; i++)
        S += min(L[2*i], L[2*i+1]);
    printf("%d\n", S);
    return 0;
}