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;
const long long mod = pow(10, 9) + 7;
const long long MOD = 998244353;
const long long INF = 1LL << 60;
const int inf = pow(10, 9) + 7;
template <class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template <class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
int divCeil(int A, int B) {return (A + (B - 1)) / B;}
long long divCeil(long long A, int B) {return (A + (B - 1)) / B;}
long long divCeil(int A, long long B) {return (A + (B - 1)) / B;}
long long divCeil(long long A, long long B) {return (A + (B - 1)) / B;}

int main()
{
  int N;
  cin >> N;
  vector<int> L(2 * N);
  for (int i = 0; i < 2 * N; i++)
  {
    cin >> L.at(N);
  }
  
  sort(L.begin(), L.end());

  int ans = 0;
  for (int i = 0; i < N; i += 2)
  {
    ans += L.at(i);
  }
  
  
  cout << ans << endl;
}