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>
#include <stdlib.h>
#include <string.h>
int size(const void *x,const void *y);
int main(){
  int N;
  int L[100];// 0 to 99
  int i = 0;
  char tmp[512];
  char *tp;
  int sum = 0;
  
  scanf("%d",&N);
  
  scanf("%s",&tmp);
  tp = strtok(tmp, "\t");
  L[i] = atoi(tp);
  i=1;
  while(tp = strtok(NULL, "\t")){
    L[i] = atoi(tp);
    i++;
  }
  //ソート
  qsort(L,i,sizeof(int),size);
  
  //
  for(int j = 0 ; j<(i/2) ;j++){
    if(L[2*j]<L[2*j +1]){
      sum += L[2*j];
    }else{
      sum += L[2*j +1];
    }
  }
  return sum;
}
 
int size(const void *x,const void *y){
  if(*(int *)x > *(int *)y){
    return 1;
  }else if(*(int *)x < *(int *)y){
    return -1;
  }else{
    return 0;
  }
}