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>
#define N 165003
int arr[N];
int main(){
	int a;
	scanf("%d",&a);
	int i;
	for(i=0;i<a*2;i++){
		scanf("%d",arr+i);
	}
	int w;
	for(i=0;i<a*2;i++){
		for(w=0;w<a*2;w++){
			if(arr[w]<arr[w+1]){
				int temp=arr[w];
				arr[w]=arr[w+1];
				arr[w+1]=temp;
			}
		}
	}
	int s=0;
	for(i=0;i<(a*2)-1;i+=2){
		s+=arr[i+1];
	}
	printf("%d",s);
}