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++ (GCC 9.2.1) (judge verdict: Accepted)

// J'aime
// Chemise Blanche

#include <bits/stdc++.h>
using namespace std;

#define int long long

#define ii pair<int,int>
#define fi first
#define sc second

#define all(x) (x).begin(),(x).end()
#define dbg(x) cerr << __LINE__ << " > " << #x << " = " << (x) << endl

void MAIN() {
  int n; cin >> n;
  vector<int> a(2 * n);
  for (auto &i : a) cin >> i;
  sort(all(a));
  int ans = 0;
  for (int i = 0; i < 2 * n; i += 2) {
    ans += min(a[i], a[i + 1]);
  }
  cout << ans << '\n';
}

signed main() {
#ifdef _DEBUG
  // freopen("in" , "r", stdin );
  // freopen("out", "w", stdout);
#endif
  ios::sync_with_stdio(0); cin.tie(0);
  int T = 1;
  // cin >> T;
  while (T--) MAIN();
}