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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
class Program
{
static int n = int.Parse(Console.ReadLine());//int.Parseは文字列を整数に変換。
static int[] parents = new int[n];//親

	static void Main()
	{
  long answer = 0;
  int[][] costs = new int[2*(n-1)][];//２頂点と距離
  int[][] vertexes = new int[n][];//ここに受け取る
    for(int i = 0; i < n; i++)
    {
    int[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),int.Parse);//１行ごとに受け取る
    vertexes[i] = new int[]{i,nums[0],nums[1]};//頂点の座標をジャグ配列へ
    }
    Array.Sort(vertexes, (a, b) => a[1] - b[1]);//x座標の昇順に並べる
    for(int i = 0; i < n-1; i++)
    {
    costs[i] = new int[]{vertexes[i][0],vertexes[i+1][0],
      vertexes[i+1][1] - vertexes[i][1]};
    }

    Array.Sort(vertexes, (a, b) => a[2] - b[2]);//y座標の昇順に並べる
    for(int i = 0; i < n-1; i++)
    {
    costs[(n-1)+i] = new int[]{vertexes[i][0],vertexes[i+1][0],
      vertexes[i+1][2] - vertexes[i][2]};
    }

    Array.Sort(costs, (a, b) => a[2] - b[2]);//コストの昇順に並べる

    for(int i = 0; i < n; i++)//個体の結合
    {
    parents[i] = i;
    }

    for(int i = 0; i < costs.Length; i++)
    {
      
        if(whatParent(costs[i][0]) != whatParent(costs[i][1]))
        { 
        
          parents[whatParent(costs[i][0])] = whatParent(costs[i][1]);//親の統一
          answer += costs[i][2];
        }
          
    }

	Console.WriteLine(answer);//WriteLineをWriteとすると、改行なしで出力。
	}

  static int whatParent(int child)
  {
    if(parents[child] == child) return child;
    else return parents[child] = whatParent(parents[child]);
  }
  
}