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.*;

class Main{
    static PriorityQueue<Edge> pq;
    static HashSet<Edge>[] G;
    static boolean[] visited;
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        G = new HashSet[N+1];
        visited = new boolean[N+1];
        for(int i=0;i<=N;i++) G[i] = new HashSet<>();
        int[][] IdXY = new int[N][3];
        for(int i=0;i<N;i++){
            int x = sc.nextInt();
            int y = sc.nextInt();
            IdXY[i] = new int[]{i,x,y};
        }
        // Prim
        pq = new PriorityQueue<>();
        myComparator1 mc1 = new myComparator1();
        Arrays.sort(IdXY,mc1);
        for(int i=0;i<N-1;i++){ // sort by x-axis
            int a = Math.min(IdXY[i][0],IdXY[i+1][0]);
            int b = Math.max(IdXY[i][0],IdXY[i+1][0]);
            G[a].add(new Edge(b,IdXY[i+1][1]-IdXY[i][1]));
            G[b].add(new Edge(a,IdXY[i+1][1]-IdXY[i][1]));

        }
        myComparator2 mc2 = new myComparator2();
        Arrays.sort(IdXY,mc2);
        for(int i=0;i<N-1;i++){ // sort by x-axis
            int a = Math.min(IdXY[i][0],IdXY[i+1][0]);
            int b = Math.max(IdXY[i][0],IdXY[i+1][0]);
            G[a].add(new Edge(b,IdXY[i+1][2]-IdXY[i][2]));
            G[b].add(new Edge(a,IdXY[i+1][2]-IdXY[i][2]));
        }
        long ans = 0;
        visit(0);
        while(pq.size()>0){
            Edge cur = pq.poll();
            if(!visited[cur.to]){
                ans += cur.length;
                visit(cur.to);
            }
        }
        System.out.println(ans);
    }
    static void visit(int i){
        visited[i]=true;
        for(Edge e:G[i]){
            if(!visited[e.to]) pq.add(e);
        }
    }
    static class Edge implements Comparable<Edge>{
        int to;
        int length;
        @Override
        public int compareTo(Edge e){
            return this.length-e.length;
        }
        Edge(int to, int cost){
            this.to=to;
            this.length=cost;
        }
    }
    static class myComparator1 implements Comparator<int[]>{
        public int compare(int[] o1, int[] o2){
            return o1[1]-o2[1];
        }
    }
    static class myComparator2 implements Comparator<int[]>{
        public int compare(int[] o1, int[] o2){
            return o1[2]-o2[2];
        }
    }
}