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 main(){
  int N;
  scanf("%d",&N);
  int L[2*N];
  int i,j,k,tmp,x=0;
  for(i=0;i<2*N;i++)
  scanf("%d ",&L[i]);
  
  for (i=0; i<2*N; ++i) {
    for (j=i+1; j<2*N; ++j) {
      if (L[i] > L[j]) {
        tmp =  L[i];
        L[i] = L[j];
        L[j] = tmp;
      }
    }
  }
  
  for(i=0;i<N;i+=2){
    x += L[i];
  }
  
  printf("%d\n",x);
  
  return 0;
}