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(void){
int n,i,j,temp;
int l[200];
int sum=0;
 
scanf("%d",&n);
for(i=0;i<n*2;i++){
scanf("%d",&l[i]);
}
 
for(i=0;i<=n*2-2;i++){
for(j=i;j<=n*2-2;j++){
if(l[j]>l[j+1]){
temp=l[j];
l[j]=l[j+1];
l[j+1]=temp;
}
}
}
 
for(i=0;i<n*2;){
sum+=l[i];
i+=2;
}
 
printf("%d",sum);
 
return(0);
}