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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 1000000000000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;


#define SIZE 100005


struct Node{
	bool operator<(const struct Node &arg) const{
		return weight < arg.weight;
	};
	int from,to,weight;
};

struct Info{
	Info(int arg_node,int arg_value){
		node = arg_node;
		value = arg_value;
	}
	bool operator<(const struct Info &arg) const{

		return value < arg.value;
	}
	int node,value;
};

int parent[SIZE],height[SIZE];

int getParent(int id){
	if(parent[id] == id)return id;
	else{
		return parent[id] = getParent(parent[id]);
	}
}

bool isSame(int a, int b){
	return getParent(a) == getParent(b);
}

void unite(int a,int b){
	int a_parent = getParent(a);
	int b_parent = getParent(b);

	if(a_parent == b_parent)return;	//元々同じグループ

	if(height[a_parent] > height[b_parent]){ //高い方に吸収する
		parent[b_parent] = a_parent;
	}else{
		if(height[a_parent] == height[b_parent])height[b_parent]++;
		parent[a_parent] = b_parent;
	}
}


int main(){

	int V,E;

	scanf("%d",&V);
	E = 2*(V-1);

	vector<Info> X,Y;
	for(int i = 0; i < V; i++){

		int x,y;
		scanf("%d %d",&x,&y);
		X.push_back(Info(i,x));
		Y.push_back(Info(i,y));
	}

	sort(X.begin(),X.end());
	sort(Y.begin(),Y.end());

	Node nodes[E];

	int index = 0;

	for(int i = 1; i < V; i++){

		nodes[index].from = X[i-1].node;
		nodes[index].to = X[i].node;
		nodes[index].weight = abs(X[i-1].value-X[i].value);
		index++;
	}

	for(int i = 1; i < V; i++){

		nodes[index].from = Y[i-1].node;
		nodes[index].to = Y[i].node;
		nodes[index].weight = abs(Y[i-1].value-Y[i].value);
		index++;
	}

	sort(nodes,nodes+E);

	//統合するのは頂点
	for(int i = 0; i < V; i++){
		parent[i] = i;
		height[i] = 0;
	}

	ll sum = 0;

	for(int i = 0; i < E; i++){
		if(!isSame(nodes[i].from,nodes[i].to)){ //同じグループでなければくっつけて、そのコストを足す
			sum += nodes[i].weight;
			unite(nodes[i].from,nodes[i].to);
		}
	}

	printf("%lld\n",sum);

	return 0;
}