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 N;
  int L[200]={0};
  int i,j,tmp;
  int sum;
  scanf("%d",&N);
  N*=2;
  for(i=0;i<N;i++){
    scanf("%d",&L[i]);
  }
  for(i=0;i<N;i++){
    for(j=i+1;j<N;j++){
      if(L[i]>L[j]){
	tmp=L[i];
	L[i]=L[j];
	L[j]=tmp;
      }
    }
  }

  sum=0;
  for(i=0;i<N;i+=2){
    sum+=L[i];
  }
  printf("%d\n",sum);

  return 0;
}