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>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int cmp(const void *a,const void *b){
    if(*(int*)a>*(int*)b){return -1;}
    else if(*(int*)a<*(int*)b){return 1;}
    else {return 0;}
}
int min(int a,int b){

    if(a>=b){return b;}
    return a;
}

int main(void){
int n;
int l[205];
int sum=0;
scanf("%d",&n);
for(int i=0;i<2*n;i++){
    scanf("%d",&l[i]);
}
qsort(l,2*n,sizeof(int),cmp);

for(int i=0;i<2*n-1;i=i+2){
    sum=sum+min(l[i],l[i+1]);
}
printf("%d\n",sum);
}