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;
#define rep(i,j,n) for(int i=(int)(j);i<(int)(n);i++)
#define REP(i,j,n) for(int i=(int)(j);i<=(int)(n);i++)
#define MOD 1000000007
#define int long long
#define ALL(a) (a).begin(),(a).end()
#define vi vector<int>
#define vii vector<vi>
#define pii pair<int,int>
#define priq priority_queue<int>
#define disup(A,key) distance(A.begin(),upper_bound(ALL(A),(int)(key)))
#define dislow(A,key) distance(A.begin(),lower_bound(ALL(A),(int)(key)))
#define tii tuple<int,int,int>
#define Priq priority_queue<int,vi,greater<int>>
#define pb push_back
#define mp make_pair
signed main(){
  int N; cin>>N;
  vi A(2*N);
  rep(i,0,2*N) cin>>A[i];
  sort(ALL(A));
  int ans=0;
  rep(i,0,N) ans+=A[i*2];
  cout<<ans<<endl;
  }