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 pb(n) push_back(n)
#define fi first
#define se second
#define all(r) (r).begin(),(r).end()
#define vmax(ary) *max_element(all(ary))
#define vmin(ary) *min_element(all(ary))
#define debug(x) cout<<#x<<": "<<x<<endl
#define fcout(n) cout<<fixed<<setprecision((n))
#define scout(n) cout<<setw(n)
#define vary(type,name,size,init) vector< type> name(size,init)
#define vvl(v,w,h,init) vector<vector<ll>> v(w,vector<ll>(h,init))
#define mp(a,b) make_pair(a,b)

#define rep(i,n) for(int i = 0; i < (int)(n);++i)
#define REP(i,a,b) for(int i = (a);i < (int)(b);++i)
#define repi(it,array) for(auto it = array.begin(),end = array.end(); it != end;++it)
#define repa(n,array) for(auto &n :(array))

using ll = long long;
using pii = pair<int,int> ;
using pll = pair<ll,ll> ;

const int mod = 1000000007;
constexpr int inf = ((1<<30)-1)*2+1 ;
constexpr double PI = acos(-1.0) ;
double eps = 1e-10 ;
const int dy[] = {-1,0,1,0,1,-1,1,-1};
const int dx[] = {0,-1,0,1,1,-1,-1,1};

inline bool value(int x,int y,int w,int h){
  return (x >= 0 && x < w && y >= 0 && y < h);
}

template<typename T>
void Unique(vector<T> &v){
  sort(all(v));
  v.erase(unique(all(v)),v.end());
}
template<typename T,typename U>
ll FindErase(T &v,U tar){
  ll cnt = 0;
  for(auto it = v.begin(); it != v.end();){
    if(*it == tar){
      it = v.erase(it);
      ++cnt;
    }
    else{
      ++it;
    }
  }
  return cnt;
}

template<typename T>
bool SuffixErase(T &v,size_t suf){
  if(suf > v.size()) return false;
  for(auto it = v.begin(); it != v.end();){
    if(distance(v.begin(),it) == suf){
      v.erase(it);
      return true;
    }
    else{
      ++it;
    }
  }
  return false;
}

template<typename T>
T ston(string& str, T n){
  istringstream sin(str) ;
  T num ;
  sin >> num ;
  return num ;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vary(int,v,2 * N,0);
  rep(i,2 * N){
    cin >> v[i];
  }
  sort(all(v));
  ll ans = 0;
  rep(i,N){
    ans += v[2 * i];
  }
  cout << ans << endl;
  return 0;
}