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>

int main(void)
{
	int N;
	scanf("%d",&N);
	
	int L[2*N];
	int i,j,tmp;
	for(i=0;i<2*N;i++)
	{
		scanf("%d",&L[i]);
	}
	for(i = 0 ; i < 2*N;i++)
	{
		for(j=i;j < 2*N-1;j++)
		{
			if(L[j] > L[j+1])
			{
				tmp = L[j];
				L[j] = L[j+1];
				L[j+1] = tmp;
			}
		}
	}
	int ans = 0;
	for(i = 0; i < 2*N;i+=2)
	{
		ans = ans+L[i];
	}
	printf("%d",ans);
	return 0;
}