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 data[101];
int N;
int r_N;
int Quick_Sort(int left,int right){
    if(left < right){
        int i = left + 1, j = right, splitting = data[left], temp;
        do{
            while(i < right && data[i] <= splitting) i++;   // Note the "<" and "<="
            while(j > left && data[j] > splitting) j--;
            if(i < j){
                temp = data[i]; data[i] = data[j]; data[j] = temp;   // Note this line, swap data[i + 1] and data[i]
            }
        }while(i < j);
        temp = data[j]; data[j] = data[left]; data[left] = temp;   // Note this line, swap data[j] and the data[left] (splitting)
        Quick_Sort(left, j - 1);
        Quick_Sort(j + 1, right);
    }
    return 0;
}

int FindAns(){
    int ans=0;
    int y=0;
    for(y=0; y<r_N; y++){
        if (y==0 || y%2==0) {
            ans= ans +data[y];
        }
    }
    
    
    
    return ans;
}
int main(void){
    
    int i=0;
    
    scanf("%d", &N);
    r_N = 2*N;
    
    for(i=0; i<r_N; i++){
        scanf("%d", &data[i]);
    }
    
    Quick_Sort(0, r_N-1);
    
    printf("%d\n",FindAns());
    
    return 0;
}