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 <iostream>
#include <cmath>
#include <vector>
#include <bitset>
#include <algorithm>
#include <stack>
#include <queue>
#include <map>
#include <climits>
#include <set>
#include <map>
#include <iomanip>
#include <cassert>
#include <functional>
#include <cstring>

using namespace std;

#define mp make_pair
#define FOR(i, a, b) for(int (i)=a;(i)<(b);++(i))
#define rep(i, n)  FOR(i,0,n)
#define rrep(i,m,n) for(int i=(int)(m); i>=(int)(n); i--)
#define vsort(v) sort((v).begin(), (v).end());					//小さい順
#define rvsort(v) sort(v.begin(), v.end(),greater<>());		//大きい順
#define YES cout<<"YES"<< endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl  
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl
#define all(c) (c).begin(),(c).end()
#define ll long long
#define lb long double
#define vi vector<int>
#define vvi vector<vector<int>>
#define vl vector<ll>
#define vvl vector<vector<ll>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define vs vector<string>
#define vc vector<char>
#define vvc vector<vector<char>>
#define Print(p) cout<<(p)<<endl
#define F first
#define S second
#define pb push_back
#define mod 1000000007LL
#define INF 1234567890123456
typedef pair<int, int> P;
typedef pair<int, P> PP;

const int maxn =1e3 + 10;

//initialization here
//
int l[201];
//
//initialization finish


// your function is here
//

//
// your function finished



int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	// your code is here
	//
	int n; cin >> n;
	

	rep(i, 2 * n) cin >> l[i];

	sort(l, l + 2*n);

	int ans = 0;
	for (int i = 0; i < 2 * n; i += 2) ans += l[i];

	Print(ans);

	//
	// your code finished


	return 0;
}