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: Runtime Error)


#include <stdio.h>
#include <string.h>
void bubblesort(int *N,int len){
    
    int i,j;
    int temp;
    for(i=0; i<len; i++){
        for(j=len-1; j>i; j--){
            if(N[j] < N[j-1]){
                temp = N[j];
                N[j] = N[j-1];
                N[j-1] = temp;
            }
        }
    }
}


int main(int argc, const char * argv[]) {
    int N=0;
    int L[100];
    int i;
    int sum=0;
    //N本設定
   scanf("%d",&N);
    //L入力
   for(i=0;i<2*N;i++){
        scanf("%d",&L[i]);
   }

    
    //ソート降順
    bubblesort(L,2*N);

    //奇数ごとにとって足す
    for(i=0;i<2*N;i+=2){
        
        sum+=L[i];
        
    }
    
    //sum表示
    printf("%d\n",sum);
    
    
    return 0;
    
}