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 <string.h>
#include <stdlib.h>
int main(void){
  int n;
  int L[201];
  int tmp, ans = 0;

  scanf("%d", &n);

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

   /* 数値を降順にソート */
  for (int i=0; i< n*2; i++) {
    for (int j=i; j<n*2; j++) {
      if (L[i] > L[j]) {
        tmp =  L[i];
        L[i] = L[j];
        L[j] = tmp;
      }
    }
  }

  for(int i = 0; i < n; i++){
    ans += L[i*2];
  }

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

  return 0;
}