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 test
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = ReadInt();
            var point = new long[N][];
            for (int i = 0; i < N; i++)
            {
                var input = ReadArrayLong();
                point[i] = new long[3] { input[0], input[1], i };
            }
            var costs = new List<long[]>();
            point = point.OrderBy(x => x[0]).ToArray();
            for (int i = 0; i < N; i++)
            {
                if (i == 0)
                {
                    costs.Add(new long[] { point[i][2], point[i + 1][2], point[i + 1][0] - point[i][0] });
                }
                else if (i == N - 1)
                {
                    costs.Add(new long[] { point[i - 1][2], point[i][2], point[i][0] - point[i - 1][0] });
                }
                else
                {
                    costs.Add(new long[] { point[i][2], point[i + 1][2], point[i + 1][0] - point[i][0] });
                    costs.Add(new long[] { point[i - 1][2], point[i][2], point[i][0] - point[i - 1][0] });
                }
            }
            point = point.OrderBy(y => y[1]).ToArray();
            for (int i = 0; i < N; i++)
            {
                if (i == 0)
                {
                    costs.Add(new long[] { point[i][2], point[i + 1][2], point[i + 1][1] - point[i][1] });
                }
                else if (i == N - 1)
                {
                    costs.Add(new long[] { point[i - 1][2], point[i][2], point[i][1] - point[i - 1][1] });
                }
                else
                {
                    costs.Add(new long[] { point[i][2], point[i + 1][2], point[i + 1][1] - point[i][1] });
                    costs.Add(new long[] { point[i - 1][2], point[i][2], point[i][1] - point[i - 1][1] });
                }
            }
            costs = costs.Distinct().OrderBy(c => c[2]).ToList();
            var uf = new UnionFind(N);
            var ans = 0L;
            foreach(var c in costs)
            {
                if (!uf.Same(c[0], c[1]))
                {
                    uf.Unite(c[0], c[1]);
                    ans += c[2];
                }
            }

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

        static string Read() { return Console.ReadLine(); }
        static int ReadInt() { return int.Parse(Read()); }
        static long ReadLong() { return long.Parse(Read()); }
        static int[] ReadArrayInt() { return Read().Split(' ').Select(s => int.Parse(s)).ToArray(); }
        static long[] ReadArrayLong() { return Read().Split(' ').Select(s => long.Parse(s)).ToArray(); }
    }

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

        public UnionFind(long n)
        {
            par = new long[n];
            rank = new long[n];
            for (var i = 0L; i < n; i++)
            {
                par[i] = i;
                rank[i] = 0;
            }
        }

        public long Root(long x)
        {
            return par[x] == x ? x : par[x] = Root(par[x]);
        }

        public bool Same(long x, long y)
        {
            return Root(x) == Root(y);
        }

        public void Unite(long x, long y)
        {
            x = Root(x);
            y = Root(y);
            if (x == y) return;

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