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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <vector>
#include <sstream>
#include <set>
using namespace std;
using ll = long long;

class range {private: struct I{int x;int operator*(){return x;}bool operator!=(I& lhs){return x<lhs.x;}void operator++(){++x;}};I i,n;
public:range(int n):i({0}),n({n}){}range(int i,int n):i({i}),n({n}){}I& begin(){return i;}I& end(){return n;}};

int main() {
  int N;
  while (cin >> N) {
    int v[2 * N];
    for (int i : range(2 * N)) cin >> v[i];
    sort(v, v + 2 * N);
    int x = 0;
    for (int i : range(N)) x += v[2 * i];
    cout << x << endl;
  }
}