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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		City[] cities = new City[n];
		for (int i = 0; i < n; i++) {
		    cities[i] = new City(i, sc.nextInt(), sc.nextInt());
		}
		Arrays.sort(cities, new Comparator<City>() {
		    public int compare(City c1, City c2) {
		        return c1.x - c2.x;
		    }
		});
		PriorityQueue<Road> queue = new PriorityQueue<>();
		for (int i = 1; i < n; i++) {
		    queue.add(new Road(cities[i].x - cities[i - 1].x, cities[i], cities[i - 1]));
		}
		Arrays.sort(cities, new Comparator<City>() {
		    public int compare(City c1, City c2) {
		        return c1.y - c2.y;
		    }
		});
		for (int i = 1; i < n; i++) {
		    queue.add(new Road(cities[i].y - cities[i - 1].y, cities[i], cities[i - 1]));
		}
		UnionFindTree uft = new UnionFindTree(n);
		long total = 0;
		while (queue.size() > 0) {
		    Road r = queue.poll();
		    if (!uft.same(r.from.id, r.to.id)) {
		        uft.unite(r.from.id, r.to.id);
		        total += r.cost;
		    }
		}
	    System.out.println(total);
   }
   
   static class UnionFindTree {
       int[] parents;
       
       public UnionFindTree(int size) {
           parents = new int[size];
           for (int i = 0; i < size; i++) {
               parents[i] = i;
           }
       }
       
       public int find(int x) {
           if (parents[x] == x) {
               return x;
           } else {
               return parents[x] = find(parents[x]);
           }
       }
       
       public boolean same(int x, int y) {
           return find(x) == find(y);
       }
       
       public void unite(int x, int y) {
           int xx = find(x);
           int yy = find(y);
           if (xx == yy) {
               return;
           }
           parents[xx] = yy;
       }
   }
   
   static class Road implements Comparable<Road> {
       int cost;
       City from;
       City to;
       
       public Road (int cost, City from, City to) {
           this.cost = cost;
           this.from = from;
           this.to = to;
       }
       
       public int compareTo(Road another) {
           return cost - another.cost;
       }
   }
   
   static class City {
       int id;
       int x;
       int y;
       
       public City(int id, int x, int y) {
           this.id = id;
           this.x = x;
           this.y = y;
       }
   }
}