Problem:
Score : 500 points

Problem Statement

There are N towns on a plane. The i-th town is located at the coordinates (x_i,y_i). There may be more than one town at the same coordinates.

You can build a road between two towns at coordinates (a,b) and (c,d) for a cost of min(|a-c|,|b-d|) yen (the currency of Japan). It is not possible to build other types of roads.

Your objective is to build roads so that it will be possible to travel between every pair of towns by traversing roads. At least how much money is necessary to achieve this?

Constraints

2 ≤ N ≤ 10^5

0 ≤ x_i,y_i ≤ 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_N y_N

Output

Print the minimum necessary amount of money in order to build roads so that it will be possible to travel between every pair of towns by traversing roads.

Sample Input 1

3
1 5
3 9
7 8

Sample Output 1

3

Build a road between Towns 1 and 2, and another between Towns 2 and 3. The total cost is 2+1=3 yen.

Sample Input 2

6
8 3
4 9
12 19
18 1
13 5
7 6

Sample Output 2

8

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

class PriorityQueue
  attr_reader :size

  def initialize(order: :asc)
    @size = 0
    @nodes = []
    @operator = (order == :asc) ? [:>, :<] : [:<, :>]
  end

  def empty?
    @size.zero?
  end

  def push(node)
    if @nodes[0].nil?
      @nodes[0] = node
    else
      index = size
      @nodes[index] = node

      loop {
        parent = (index-1)/2

        if (@nodes[parent].last <=> node.last).public_send(@operator.first, 0)
          @nodes[index], @nodes[parent] = @nodes[parent], @nodes[index]
          index = parent

          break if index.zero?
        else
          break
        end
      }
    end

    @size += 1
  end

  alias_method :<<, :push

  def pop
    node = @nodes.first
    @nodes[0], @nodes[size - 1] = @nodes[size - 1], nil
    index = 0

    loop {
      left = 2*index+1
      right = 2*index+2

      if @nodes[left].nil? && @nodes[right].nil?
        break
      elsif @nodes[right].nil? && @nodes[left]
        if (@nodes[left].last <=> @nodes[index].last).public_send(@operator.last, 0)
          @nodes[index], @nodes[left] = @nodes[left], @nodes[index]
          index = left
        else
          break
        end
      elsif @nodes[left].nil? && @nodes[right]
        if (@nodes[right].last <=> @nodes[index].last).public_send(@operator.last, 0)
          @nodes[index], @nodes[right] = @nodes[right], @nodes[index]
          index = right
        else
          break
        end
      elsif (@nodes[left].last <=> @nodes[right].last).public_send(@operator.last, 0)
        @nodes[index], @nodes[left] = @nodes[left], @nodes[index]
        index = left
      else
        @nodes[index], @nodes[right] = @nodes[right], @nodes[index]
        index = right
      end
    }

    @size -= 1
    node.first
  end

  def top
    @nodes.first
  end
end

N = gets.to_i

city = []

N.times do
  city << gets.chomp.split.map(&:to_i)
end

pque = PriorityQueue.new

0.upto(N - 2) do |i|
  (i + 1).upto(N - 1) do |j|
    c1 = city[i]
    c2 = city[j]
    cost = [(c1[0] - c2[0]).abs, (c1[1] - c2[1]).abs].min
    pque.push([[i, j, cost], cost])
  end
end

check_list = {}
total_cost = 0
cnt = 0

while cnt < N
  i, j, cost = pque.pop

  unless check_list[i] && check_list[j]
    unless check_list[i]
      check_list[i] = true
      cnt += 1
    end

    unless check_list[j]
      check_list[j] = true
      cnt += 1
    end

    total_cost += cost
  end
end

puts total_cost