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 <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <map>
#include <memory>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <typeinfo>
#include <utility>
#include <vector>
#include <array>
#include <chrono>
#include <random>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#define INIT std::ios::sync_with_stdio(false);std::cin.tie(0);
// VAR(int, x);
#define VAR(type, ...)type __VA_ARGS__;Scan(__VA_ARGS__);
template<typename T> void Scan(T& t) { std::cin >> t; }
template<typename First, typename...Rest>void Scan(First& first, Rest&...rest) { std::cin >> first; Scan(rest...); }
#define OUT(d) std::cout<<d;
#define FOUT(n, d) std::cout<<std::fixed<<std::setprecision(n)<<d;
#define SOUT(n, c, d) std::cout<<std::setw(n)<<std::setfill(c)<<d;
#define SP std::cout<<" ";
#define TAB std::cout<<"\t";
#define BR std::cout<<"\n";
#define ENDL std::cout<<std::endl;
#define FLUSH std::cout<<std::flush;
#define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i;
#define MAT(type, c, m, n) std::vector<std::vector<type>> c(m, std::vector<type>(n));for(auto& r:c)for(auto& i:r)std::cin>>i;
#define ALL(a) (a).begin(),(a).end()
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define RFOR(i, a, b) for(int i=(b)-1;i>=(a);--i)
#define REP(i, n) for(int i=0;i<int(n);++i)
#define RREP(i, n) for(int i=(n)-1;i>=0;--i)
#define PAIR std::pair<int, int>
#define IN(a, x, b) (a<=x && x<=b)
#define IN2(a0, y, a1, b0, x, b1) (a0<=y && y<a1 && b0<=x && x<b1)
#define SHOW(d) {std::cout << #d << "\t:" << d << "\n";}
#define SHOWVECTOR(v) {std::cout << #v << "\t:";for(const auto& i : v){std::cout << i << " ";}std::cout << "\n";}
#define SHOWVECTOR2(v) {std::cout << #v << "\t:\n";for(const auto& i : v){for(const auto& j : i){std::cout << j << " ";}std::cout << "\n";}}
#define SHOWPAIRVECTOR2(v) {std::cout << #v << "\t:\n";for(const auto& i : v){for(const auto& j : i){std::cout<<'('<<j.first<<", "<<j.second<<") ";}std::cout << "\n";}}
#define SHOWPAIRVECTOR(v) {for(const auto& i:v){std::cout<<'('<<i.first<<", "<<i.second<<") ";}std::cout<<"\n";}
#define CHECKTIME(state) {auto start=std::chrono::system_clock::now();state;auto end=std::chrono::system_clock::now();auto res=std::chrono::duration_cast<std::chrono::nanoseconds>(end-start).count();std::cerr<<"[Time:"<<res<<"ns  ("<<res/(1.0e9)<<"s)]\n";}

#define int ll
using ll = long long;
constexpr int INFINT = 1 << 30;
constexpr ll INFLL = 1LL << 60;
constexpr double EPS = 0.0000001;
constexpr int MOD = 1000000007;

signed main() {
	INIT;
	VAR(int, n);
	VEC(int, l, n*2);
	std::sort(ALL(l));
	int sum = 0;
	REP(i, n) {
		sum += l[i * 2];
	}
	OUT(sum)BR;
	return 0;
}