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>
#include <ext/pb_ds/assoc_container.hpp>

#define x first
#define y second
#define ndl '\n'
#define mp make_pair
#define up_b upper_bound
#define low_b lower_bound
#define sz(x) (int)x.size()
#define bit __builtin_popcount
#define all(x) x.begin(),x.end()
#define show(x) cout<<(#x)<<" is "<<(x)<<endl

using namespace std;
using namespace __gnu_pbds;

template<typename T> using indexed_set=tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
typedef pair<int,ll> pil;
typedef pair<ll,int> pli;
typedef vector<vector<int>> matrix;

const ll INF=1e18+123;
const ld EPS=1e-9;
const int inf=1e9+123;
const int MOD=1e9+7;
const int N=5e5+123;
const int M=1e6+123;
const int dx[]={0,0,1,-1};
const int dy[]={1,-1,0,0};

int a[N];

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin>>n;
    n+=n;
    for(int i=1;i<=n;i++)cin>>a[i];
    sort(a+1,a+1+n);
    int ans=0;
    for(int i=1;i<=n;i+=2)ans+=a[i];
    cout<<ans;
    return 0;
}