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<stdlib.h>
#include<string.h>
#include<math.h>
 
#define INF 1000000000
 
#define min(a,b) (((a)<(b))?(a):(b))
#define max(a,b) (((a)>(b))?(a):(b))

long N;
long L[101];

int comp(const void *x, const void *y) {
	const long* a=(const long*)x;
	const long* b=(const long*)y;
	if(*a>*b)return 1;
	if(*a<*b)return -1;
	return 0;
}

int main() {
	long i, j;
	long ans = 0;
	
	scanf("%ld", &N);
	for(i=0;i<2*N;i++) {
		scanf("%ld", &L[i]);
	}
	
	qsort((void *)L, 2*N, sizeof(long), comp);
	
	for(i=0;i<2*N;i+=2) {
		ans += L[i];
	}
	
	printf("%ld\n", ans);
	
	return 0;
}