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>
#define nil 1000
void sort(int A[]){
    int i=0;
    while(A[i]!=nil){
        int j=i+1,minx=i;
        while(A[j]!=nil){
            if(A[j]<A[minx]){
                minx=j;
            }
            j++;
        }
        int s;
        s=A[i];
        A[i]=A[minx];
        A[minx]=s;
        i++;
    }
    
}

int main(){
    int n;
    scanf("%d",&n);
    int L[2*n+1];
    int i;
    for(i=0;i<=2*n-1;i++)scanf("%d",&L[i]);
    L[2*n]=nil;
    sort(L);
    int ans=0;
    for(i=0;i<=n-1;i++){
        ans+=L[2*i];
    }
    printf("%d",ans);
    return 0;
}