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 <math.h>

int main(){
  
  int i,n,j3;
  int a[200];
  
  scanf("%d",&n);
  j3=2*n;
  
  for(i=0;i<j3;i++){
    a[i]=0;
    scanf("%d",&a[i]);
  }

  int i3,i4;
  int alpha;
  //---narabekae //bubble sort
  for(i3=0;i3<(j3-1);i3++){
    for(i4=(j3-1);i4>i3;i4=i4-1){
      if(a[i4-1]>a[i4]){
        alpha=a[i4-1];
        a[i4-1]=a[i4];
        a[i4]=alpha;
      }      
    }
  }
  
  int count;
  count=0;
  for(i=0;i<n;i++){
    count=count+a[2*i];
              //a[2*i+1];
  }
  printf("%d",count);

  return 0;
}