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>

int main(void){

  int a , i, j, hoge, sum = 0;
  scanf("%d",&a);
  a = a * 2;

  int x[a];

  for(i = 0; i < a; i++){
    scanf("%d",&x[i]);
  }

  for(i = 0; i < a - 1 ; i++){
    for(j = a - 1; j > i; j--){
      if(x[j] < x[j-1]){
        hoge = x[j];
        x[j] = x[j-1];
        x[j-1] = hoge;
      }
    }
  }
    for(i = 0; i < a; i++){
  //    printf("%d",x[i]);
      if(i % 2== 0){
        sum += x[i];
      }
  }
  printf("%d\n",sum);
  return 0;
  }