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(void){
    int N, L[200];
    int ans = 0;
    scanf("%d", &N);
    //L[i]を昇順ソートしながら格納
    for(int i = 0; i < 2*N; i ++){
        int num;
        scanf("%d", &num);
        int place = 0;
        if (i > 0){
            while(place < i){
                if (num < L[place]) break;
                place ++;
            }
            for(int k = i; k >= place; k --){
                L[k+1] = L[k];
            }
        }
        L[place] = num;
    }
    for(int i = 0; i < 2*N; i += 2){
        ans += L[i];
    }
    printf("%d\n", ans);    
    return 0;
}