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>

void Qsort(int x[], int left, int right);
void Swap(int x[], int i, int j);

int main(void) 
{
	int N, L[100], sum = 0;

	scanf("%d", &N);

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

	Qsort(L, 0, 2*N - 1);

	for (int i = 0; i < N; i++) {
		sum += L[2 * i];
	}

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


	return 0;
}

void Swap(int x[], int i, int j) {
	int w;

	w = x[i];
	x[i] = x[j];
	x[j] = w;

}

void Qsort(int x[], int left, int right) {

	int i, j, pivot;

	i = left;
	j = right;
	pivot = x[(i + j) / 2];

	while (1) {

		while (x[i] < pivot)
			i++;

		while (x[j] > pivot)
			j--;
		if (i >= j)
			break;

		Swap(x, i, j);
		i++;
		j--;
	}

	if (left < i - 1)
		Qsort(x, left, i - 1);
	if (right > j + 1)
		Qsort(x, j + 1, right);
}