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>
using namespace std;

using int32 = int_fast32_t;
using int64 = int_fast64_t;

const int32 INF = 1e9;
const int32 MOD = 1e9+7;
const int64 LLINF = 1e18;

#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define ANS(n) cout << (n) << endl
#define REP(i,n) for(int64 i=0;i<(n);++i)
#define FOR(i,a,b) for(int64 i=(a);i<(b);i++)
#define FORR(i,a,b) for(int64 i=(a);i>=(b);i--)
#define ALL(obj) (obj).begin(),(obj).end()
#define pii pair<int32,int32>
#define pll pair<int64,int64>
#define pb(a) push_back(a)
#define mp make_pair

int main()
{ int N, ans = 0;
   cin >> N;
  std::vector<int> L(2*N);
  for(int i = 0; i < 2*N; i++){
      cin >> L[i];
  }
  sort(L.begin(),L.end());
  for(int i = 0; i < 2*N; i += 2){
     ans += L[i]; 
  }
  cout << ans;
    
    return 0;
}