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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <tuple>
using namespace std;
const int mod = 1000000007;

int main() {

#ifndef	ONLINE_JUDGE
	freopen("test.txt", "r", stdin);
	freopen("result.txt", "w", stdout);
#endif
	/*
	int testcase;
	cin >> testcase;
	for (int index = 1; index <= testcase; ++index) {
		cout << "Case #" << index << ": ";*/
		
	int n;
	cin >> n;
	vector<int> nums(2 * n);
	for (int i = 0; i < 2 * n; ++i) {
		cin >> nums[i];
	}
	sort(nums.begin(), nums.end());
	int result = 0;
	for (int i = 2 * n - 1; i >= 1; i = i - 2) {
		result += min(nums[i], nums[i - 1]);
	}
	cout << result << endl;
		
	
#ifndef ONLINE_JUDGE
	fclose(stdin);
	fclose(stdout);
	system("result.txt");
#endif
}