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)

# define _CRT_SECURE_NO_WARNINGS 1
# include <iostream>
# include <string>
# include <bitset>
# include <vector>
# include <algorithm>
# include <cstdlib>
# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iomanip>
# include <queue>
# include <sstream>
# include <climits>
# include <cmath>
# include <list>
# include <functional>
# include <string>
# include <ctime>
# include <set>
# include <map>
# include <stack>
using namespace std;
# define M_PI 3.141592
# define FOR(i,n) for(int i=0;i<(int)n;i++)
# define FORI(i,k,n) for(int i=k;i<(int)n;i++)
# define toRad 2.0*M_PI/360.0
# define inin(x) int x;cin>>x;
# define all(x) x.begin(),x.end()
# define debug(x) cout<<#x<<" :"<<x<<endl;
# define rep(i,n) for(int i=0;i<(int)n;i++)
# define EPS 1e-12
# define CHECK(i,a) FOR(i,a.size())cout<<#a<<"["<<i<<"] : "<<a[i]<<endl; 

int main()
{
	int n;
	int a[200];
	cin >> n;
	for (int i = 0; i < 2 * n; i++)
	{
		cin >> a[i];
	}
	sort(begin(a), begin(a)+2*n);
	int ans = 0;
	for (int i = 0; i < 2 * n; i += 2)
	{
		ans += a[i];
	}
	cout << ans << endl;
}