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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
///*
void sort_N(int *x,int n);

int main() {

	int i, n, *ary;
	int counter=0;

	scanf("%d", &n);

	ary = (int*)calloc(2 * n, sizeof(int));
	if (ary == NULL) {
		return 0;
	}

	for (i = 0; i < 2 * n; i++) {
		scanf("%d", &ary[i]);
	}

	sort_N(ary, 2 * n);

	for (i = 0; i < n; i++) {
		counter += ary[2 * i];
	}

	for (i = 0; i < 2 * n; i++) {
		printf("%d ", ary[i]);
	}
	printf("\n");

	printf("%d\n", counter);

	free(ary);

	return 0;
}

void sort_N(int *x,int n) {

	int i,j,fake;
	
	for (i = 0; i < n - 1; i++) {
		for (j = 0; j < n - 1;j++){
			if (x[j] > x[j + 1]) {
				fake = x[j];
				x[j] = x[j + 1];
				x[j + 1] = fake;
			}
		}
	}
}
//*/