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: Accepted)

class UnionFind
  def initialize(n)
    @par = (0..n).to_a
    @rank = Array.new(n, 0)
  end
  
  def find(x)
    if @par[x] == x then
      return x
    else
      @par[x] = self.find(@par[x])
      return @par[x]
    end
  end
  
  def unite(x, y)
    x = self.find(x)
    y = self.find(y)
    if x == y then
      return
    end
    if @rank[x] < @rank[y] then
      @par[x] = y
    else
      @par[y] = x
      if @rank[x] == @rank[y] then
        @rank[x] += 1
      end
    end
  end

  def same(x, y)
    return self.find(x) == self.find(y)
  end 
end

class Node
  attr_accessor :x, :y, :i
  def initialize(x, y, i)
    @x = x
    @y = y
    @i = i
  end
end

class Edge
  attr_accessor :u, :v, :cost
  def initialize(u, v, cost)
    @u = u
    @v = v
    @cost = cost
  end
end

def kruscal(edge, n)
  edge.sort_by!{|x| x.cost}
  uf = UnionFind.new(n)
  res = 0
  edge.size.times do |i| 
    e = edge[i]
    if !uf.same(e.u, e.v) then
      uf.unite(e.u, e.v)
      res += e.cost
    end
  end
  return res
end

n = gets.chomp.to_i
edge = []
nodes = []

n.times do |i|
  x, y = gets.chomp.split(" ").map(&:to_i)
  nodes << Node.new(x, y, i)
end

nx = nodes.sort_by{|x| x.x}

(n-1).times do |i|
  n1 = nx[i]
  n2 = nx[i+1]
  edge << Edge.new(n1.i, n2.i, (n1.x - n2.x).abs)
end

ny = nodes.sort_by{|x| x.y}
(n-1).times do |i|
  n1 = ny[i]
  n2 = ny[i+1]
  edge << Edge.new(n1.i, n2.i, (n1.y - n2.y).abs)
end

puts kruscal(edge, n)