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;

typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef long double LD;
typedef pair<LD, LD> PLDLD;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define CLR(a) memset((a), 0 ,sizeof(a))
#define ALL(a) a.begin(),a.end()
LL MOD = 1e9+7;
LL INF = 1e9;
double EPS=1.0e-9;


int main()
{
    int n;
    cin>>n;
    vector<int> l(n*2);
    REP(i,n*2)
        cin>>l[i];
    
    sort(ALL(l));
    int sum=0;
    REP(i,n)
    sum+=l[i*2];
    
    cout<<sum<<endl;
}