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>

int main(){
	int N,tmp,cnt;
	int i,p,q;

	scanf("%d", &N);
	
        if(N<0 && N>100) return -1;

        int L[N*2];
	for( i=0; i<N*2; i++ ){
		scanf("%d ",&L[i]);
     if (L[i] < 0 || L[i] > 100 ){
      return -1;
    }
	}

	//降順ソート
	for( p = 0; p <( N*2)-1; p++){
		for(q = p + 1; q < (N*2); q++){
			if(L[p] < L[q]){
				tmp = L[p];
				L[p] = L[q];
				L[q] = tmp;
			}
		}
	}

	//最大数の和を求める
	for( i= 0; i<N*2; i+=2 ){
		cnt += L[i+1];
	}

	printf("%d",cnt);

	return 0;
}