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;
using System.Collections.Generic;
using System.Linq;

namespace ABC065D
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = int.Parse(Console.ReadLine());

            memo[] x = new memo[N];
            memo[] y = new memo[N];
            for (int i = 0; i < N; i++)
            {
                var input = ReadLong();
                x[i] = new memo() { town = i, point = input[0] };
                y[i] = new memo() { town = i, point = input[1] };
            }

            //ソート
            x = x.OrderBy(a => a.point).ToArray();
            y = y.OrderBy(a => a.point).ToArray();

            Edge[] edgesx = new Edge[N - 1];
            Edge[] edgesy = new Edge[N - 1];
            for (int i = 0; i < N - 1; i++)
            {
                edgesx[i] = new Edge() { s = x[i].town, t = x[i + 1].town, cost = x[i + 1].point - x[i].point };
                edgesy[i] = new Edge() { s = y[i].town, t = y[i + 1].town, cost = y[i + 1].point - y[i].point };
            }

            Edge[] edges = edgesx.Concat(edgesy).ToArray();
            edges = edges.OrderBy(a => a.cost).ToArray();

            UnionFind unionFind = new UnionFind(N);

            long ans = 0;   //最小コスト
            foreach (var e in edges)
            {
                //結合済みでないなら追加
                if (!unionFind.Same(e.s, e.t))
                {
                    unionFind.Unite(e.s, e.t);
                    ans += e.cost;
                }
            }

            Console.WriteLine(ans);
            Console.ReadKey();
        }


        struct Edge
        {
            public int s;
            public int t;
            public long cost;
        }

        struct memo
        {
            public int town;
            public long point;
        }

        static int[] ReadInt()
        {
            int[] ret = Console.ReadLine().Split().Select(int.Parse).ToArray();
            return ret;
        }
        static long[] ReadLong()
        {
            long[] ret = Console.ReadLine().Split().Select(long.Parse).ToArray();
            return ret;
        }
    }


    class UnionFind
    {
        int[] par;
        int[] rank;

        public UnionFind(int n)
        {
            init(n);
        }

        void init(int n)
        {
            par = new int[n];
            rank = new int[n];

            for (int i = 0; i < n; i++)
            {
                par[i] = i;
                rank[i] = 0;
            }
        }

        public int Find(int x)
        {
            if (par[x] == x)
            {
                return x;
            }
            else
            {
                return par[x] = Find(par[x]);
            }
        }

        public void Unite(int x, int y)
        {
            x = Find(x);
            y = Find(y);
            if (x == y) return;

            if (rank[x] < rank[y])
            {
                par[x] = y;
            }
            else
            {
                par[y] = x;
                if (rank[x] == rank[y]) rank[x]++;
            }

        }
        public bool Same(int x, int y)
        {
            return Find(x) == Find(y);
        }
    }
}