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: Runtime Error)

class UnionFind
  def initialize(n)
    @size = Array.new(n + 1, 1)
    @parent = []

    n.times do |i|
      @parent[i] = i
    end
  end

  def find(x)
    if @parent[x] == x
      x
    else
      @parent[x] = find(@parent[x])
    end
  end

  def unite(x, y)
    x = find(x)
    y = find(y)
    return if x == y

    @parent[y] = x
    @size[x] += @size[y]
  end

  def same?(x, y)
    find(x) == find(y)
  end

  def size(x)
    @size[find(x)]
  end
end

N = gets.to_i
cities = Hash.new { |h, k| h[k] = Hash.new { |h, k| h[k] = [] } }
coords = N.times.map.with_index(1) do |i|
  coord = gets.split.map(&:to_i)

  cities[coord[0]][coord[1]] << i

  coord
end

x_order = coords.sort_by { |x, _y| x }
y_order = coords.sort_by { |_x, y| y }

edges = []

x_order.each_cons(2) do |(x1, y1), (x2, y2)|
  cost = x2 - x1

  cities[x1][y1].each do |c1|
    cities[x2][y2].each do |c2|
      next if c1 == c2

      edges << [c1, c2, cost]
    end
  end
end

y_order.each_cons(2) do |(x1, y1), (x2, y2)|
  cost = y2 - y1

  cities[x1][y1].each do |c1|
    cities[x2][y2].each do |c2|
      next if c1 == c2

      edges << [c1, c2, cost]
    end
  end
end

edges.sort_by! { |_from, _to, cost| cost }
uf = UnionFind.new(N)

ans = 0

edges.each do |from, to, cost|
  unless uf.same?(from, to)
    uf.unite(from, to)
    ans += cost
  end
end

puts ans