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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<iostream>
#include<cmath>
#include<vector>
#include<string>
#include<algorithm>

#define ll long long
#define LL long long
#define ULL unsigned long long 
#define ull unsigned long long 
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define RFOR(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define REP(i,n) for(int i=0;i<n;i++)
#define RREP(i,n) for(int i=(n)-1;i>=0;i--)

using namespace std;

int main()
{
	int len[200];
	int N;
	int ans=0;
	cin >> N;
	REP(i, 2*N) {
		cin >> len[i];
	}
	sort(len, len + 2 * N);
	for (int i = 0; i < N*2; i = i + 2) {
		ans +=min(len[i],len[i+1]);
	}

	cout << ans << endl;
	

	return 0;
}