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>

int main(){
    int n, i, j, k, work,sum=0;
    int kushi[200]={};

    scanf("%d",&n);

    

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

    for(i=0;i<n*2;i++){
        for(j=i+1;j<n*2;j++){
            if(kushi[i]<kushi[j]){
               work = kushi[i];
               kushi[i] = kushi[j];
               kushi[j] = work; 
            }
        }
    }

    for(k=1;k<n*2;k+=2){
        sum += kushi[k];
    }

    printf("%d",sum);

    return 0;
}