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: Runtime Error)

#include<stdio.h>

int main(void)
{
  int num;
  int length[100] = {};
  int i, j;
  int temp;
  int sum=0;

  scanf("%d", &num);

  for (i=0;i<num*2;i++){
    scanf("%d", &length[i]);
  }

  for(i=0;i<num*2;i++){
    for(j=i+1;j<num*2;j++){
      if(length[i] < length[j]){
	temp = length[i];
	length[i] = length[j];
	length[j] = temp;
      }
    }
  }
  sum = length[1];
  for(i=1;i<num*2;i+=2){
    sum += length[i+2];
    if (i+2 == num*2-1)
      break;
  }

  printf("%d\n", sum);

  return 0;
}