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];
    for(int i=0;i<2*N;i++){
        scanf("%d", &L[i]);
    }
    int M=0;
    for(int i=0;i<2*N-1;i++){
        for(int j=0;j<2*N-i;j++){
            if(L[j]>L[j+1]){
                int z;
                z = L[j+1];
                L[j+1] = L[j];
                L[j] = z;
            }
        }
    }
    for(int i=0; i<N; i++){
        M += L[2*i];
    }
    for(int i=0;i<2*N;i++){
        printf("%d ", L[i]);
    }
    printf("\n");
    printf("%d\n", M);
    return 0;
}