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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
void swap(int* a, int* b)
{
    int t = *a;
    *a = *b;
    *b = t;
}
int partition (int arr[], int low, int high)
{
    int pivot = arr[high];  
    int i = (low - 1),j;  
    for (j = low; j <= high- 1; j++)
    {
        if (arr[j] <= pivot)
        {
            i++;    
            swap(&arr[i], &arr[j]);
        }
    }
    swap(&arr[i + 1], &arr[high]);
    return (i + 1);
}
void quicksort(int a[200],int low,int high)
{
	if (low < high)
    {
        int pi = partition(a, low, high);
        quicksort(a, low, pi - 1);
        quicksort(a, pi + 1, high);
    }
}
void main()
{
	int n,i,a[200],sum=0;
	scanf("%d",&n);
	for(i=0;i<2*n;i++)
		scanf("%d",&a[i]);
	quicksort(a,0,2*n-1);
	for(i=0;i<2*n;i+=2)
		sum+=a[i];
	printf("%d\n",sum);
}