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 <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
using namespace std;
typedef long long  ll;
#define rep(i,s,n)for(ll i=s;i<n;i++)
#define repe(i,s,n)for(ll i=s;i<=n;i++)

static const double PI = 3.14159265358979;
static const ll MOD = 1e9 + 7;
static const ll INF = (ll)1 << 62;

ll l[201] = {};
int main() {
	ll n; cin >> n;
	rep(i, 0, 2 * n)cin >> l[i];

	sort(l, l + 2 * n);
	reverse(l, l + 2 * n);

	ll ans = 0;
	for (int i = 0; i < 2 * n; i += 2) {
		ans += l[i + 1];
	}
	cout << ans << endl;

	return 0;
}