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>

void main()
{int i,a[200],length,result,error,n2;
error=scanf("%d",&i);
n2=i*2;
for(i=0;i<n2;i++) {error=scanf("%d",&a[i]);
                         }

for(int i=0;i<n2;i++)
	{
      	for(int j=n2-1;j>i;j--)
		{
		if(a[j]<a[j-1])
			{
			  int t=a[j];
			  a[j]=a[j-1];
			  a[j-1]=t;
			}
	      }
	}

for(i=0;i<n2;i=i+2) {result=result+a[i];
                         }
printf("%d ",result);


}