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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
#define popCnt(x) (__builtin_popcountll(x))
typedef unsigned long long Long;

int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#ifdef Local
  freopen("test.in", "r", stdin);
#else
#define endl '\n'
#endif

  int n;
  cin >> n;
  n *= 2;
  vector<int> v(n);
  for (int& x : v) {
    cin >> x;
  }

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

  int res = 0;
  for (int i = 0; i < v.size(); i += 2) {
    res += v[i];
  }

  cout << res;
}