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>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define all(obj) (obj).begin(),(obj).end() 
#define dump(x)  cout << #x << " = " << (x) << endl;
typedef long long lint;
using namespace std;
int main(){
	int n,ans=0;
	cin>>n;
	vector<int>l(2*n);
	rep(i,2*n)cin>>l.at(i);
	sort(all(l));
	for(int i=0;i<2*n;i+=2)ans+=l.at(i);
	cout<<ans<<endl;
}