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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const double pi = acos(-1);
#define FOR(i,a,b) for (ll i=(a),__last_##i=(b);i<__last_##i;i++)
#define RFOR(i,a,b) for (ll i=(b)-1,__last_##i=(a);i>=__last_##i;i--)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define __GET_MACRO3(_1, _2, _3, NAME, ...) NAME
#define rep(...) __GET_MACRO3(__VA_ARGS__, FOR, REP)(__VA_ARGS__)
#define rrep(...) __GET_MACRO3(__VA_ARGS__, RFOR, RREP)(__VA_ARGS__)
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v) {
REP(i,v.size()){if(i)os<<" ";os<<v[i];}return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v) {
REP(i,v.size()){if(i)os<<endl;os<<v[i];}return os;}


int main()
{
	int N;
	cin >> N;
	vector<int> stick(N);

	rep(i, N) cin >> stick[i];

	vector<int> sum(N / 2);

	sort(stick.begin(), stick.end());

	

	rep(i, N / 2){ sum[i] = stick[2 * i];}

	int price_sum = 0;

	rep(i, N / 2){
		price_sum = sum[i];
	}


	cout << price_sum << endl;
}