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 Ruby (2.3.3) (judge verdict: Accepted)

#! ruby
# Try AtCoder
# author: Leonardone @ NEETSDKASU
############################################################
def gs() gets.chomp end
def gi() gets.to_i end
def gf() gets.to_f end
def gss() gs.split end
def gis() gss.map(&:to_i) end
def nmapf(n,f) n.times.map{ __send__ f } end
def ngs(n) nmapf n,:gs end
def ngi(n) nmapf n,:gi end
def ngss(n) nmapf n,:gss end
def ngis(n) nmapf n,:gis end
def arr2d(h,w,v=0) h.times.map{[v] * w} end
def for2p(hr,wr,&pr) hr.each{|i|wr.each{|j| yield(i,j)}} end
def nsum(n) n * (n + 1) / 2 end
def vcount(d,r=Hash.new(0)) d.inject(r){|r,e| r[e]+=1;r} end
############################################################

n = gi
l = gis.sort.reverse

a = 0

n.times do |i|
    a += l[i * 2 + 1]
end

puts a