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: Time Limit Exceeded)

Edge = Struct.new(:from, :to, :cost)
 
class PriorityQueue
  attr_reader :elements
  
  def initialize(e=[])
    @elements = e
  end
  
  def push(n)
    i = @elements.size
    while i > 0
      p = (i-1) / 2
      break if @elements[p].cost <= n.cost
      @elements[i] = @elements[p]
      i = p
    end
    @elements[i] = n
  end
  alias :<< :push
  
  def pop
    min = @elements[0]
    n   = @elements.pop
    unless @elements.empty?
      i = 0
      while i * 2 + 1 < @elements.size
        if @elements[i * 2 + 2].nil? or @elements[i * 2 + 1].cost < @elements[i * 2 + 2].cost
          buf = i * 2 + 1
        else
          buf = i * 2 + 2
        end
        break if @elements[buf].cost >= n.cost
        @elements[i] = @elements[buf]
        i = buf
      end
      @elements[i] = n
    end
    min
  end
  
  def empty?
    @elements.empty?
  end
end

class UnionFindTree
  def initialize(n)
    @parent = (0...n).to_a
    @rank = Array.new(n, 0)
  end
     
  def find(x)
    return x if @parent[x] == x
    return @parent[x] = find(@parent[x])
  end
     
  def unite(a, b)
    a = find(a)
    b = find(b)
    return if a == b
    if @rank[a] < @rank[b]
      @parent[a] = b
    else
      @parent[b] = a
      @rank[a] += 1 if @rank[a] == @rank[b]
    end
  end
  
  def same?(a, b)
    find(a) == find(b)
  end
end

N = gets.to_i
xy = Array.new(N) { |i| x, y = gets.split.map(&:to_i); [x, y, i] }

queue = PriorityQueue.new

xy_x = xy.sort_by{|a| [a[0]] }
xy_x.zip(xy_x.drop(1)).take(N-1).each do |(x1, y1, i1), (x2, y2, i2)|
  queue << Edge.new(i1, i2, [(x1-x2).abs, (y1-y2).abs].min)
end

xy_y = xy.sort_by{|a| [a[1]] }
xy_y.zip(xy_y.drop(1)).take(N-1).each do |(x1, y1, i1), (x2, y2, i2)|
  queue << Edge.new(i1, i2, [(x1-x2).abs, (y1-y2).abs].min)
end

tree = UnionFindTree.new(N)
n = 0
sum = 0
until queue.empty? or n == N
  e = queue.pop
  
  unless tree.same?(e.from, e.to)
    tree.unite(e.from, e.to)
    n += 1
    sum += e.cost
  end
end
puts sum