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: Runtime Error)

#include<bits/stdc++.h>
#define fr first
#define sc second
#define pb push_back
#define maxheap priority_queue<int>
#define minheap priority_queue<int, vector<int>, greater<int> >
#define ll long long
const int INF = 1e8;
const int MAXN = 3e5 + 200;
const  long double eps = 1e-9;
const long double pi = acos(-1.0);
using namespace std;
const ll MOD = 998244353;


void solve(){
    ll n, x;
    cin >> n >> x;
    if (x * 2 == n){
        cout << 3 * x << endl;
        return;
    }
    ll g = __gcd(n - 2 * x, x);
   if (x >= n - 2 * x){
       cout << n + x + x + (x  + g) << endl;
       return;
   }
   cout << 1 / 0 << endl;
   cout << n + x + 2 * ((n - 2 * x) / g);
}

int main(){
    ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
    cout.setf(ios::fixed), cout.precision(17);
    //freopen("input.in", "r", stdin);
    //freopen("output.out", "w", stdout)
    int step;
    step = 1;
    for (int i = 1; i <= step; i++)
        solve();

}