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>
#include <string.h>
#include <math.h>
#include <time.h>

#define _CRT_SECURE_NO_WARNINGS
#define TLong long long
#define TBMod 1000000007

int comp(const int *a,const int *b){
	return (*b - *a);
}

int main(int argc, char const *argv[])
{
	int n,*l;
	int cnt = 0;
	scanf("%d",&n);
	l = (int *)malloc(sizeof(int) * (2 * n));
	for (int i = 0; i < 2 * n; ++i)	scanf("%d",&l[i]);
	qsort(l,(2 * n),sizeof(int),comp);
	for (int i = 0; i < 2 * n; i += 2)
	{
		if(l[i] < l[i + 1])	cnt += l[i];
		else cnt += l[i + 1];
	}
	printf("%d\n", cnt);
	return 0;
}