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 <string.h>
#include <stdlib.h>

typedef unsigned long long ull;
typedef   signed long long sll;

#define N_MAX 100

int comp (const void *x, const void *y) {
	return *(int*)x - *(int*)y;
}

int min (int x, int y) {
	return (x < y) ? x : y;
}

int solve (void) {
	return 0;
}

int main (void) {
	int l[N_MAX * 2];
	int n, i;
	int answer;
	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d%d", &l[i*2], &l[i*2+1]);
	}
	qsort(l, n * 2, sizeof(int), comp);
	answer = 0;
	for (i = 0; i < n * 2; i += 2) answer += l[i];
	printf("%d\n", answer);
	return 0;
}