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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

from operator import itemgetter
from heapq import heappush,heappop
N = int(input())
table=[]
for i in range(N):
    a,b = map(int,input().split())
    table.append([a,b,i])
A = sorted(table,key=itemgetter(0))
B = sorted(table,key=itemgetter(1))

graph=[]
for k in range(N):#距離、点
    graph.append([min(abs(A[k][0]-A[k-1][0]),abs( A[k][1]-A[k-1][1])),A[k][2],A[k-1][2]])
for k in range(N):
    graph.append([min(abs(B[k][0]-B[k-1][0]),abs( B[k][1]-B[k-1][1])),B[k][2],B[k-1][2]])

class Unionfindtree:
    def __init__(self,number):
        self.par = [i for i in range(number)]
        self.rank = [0]*(number)

    def find(self,x):#親を探す xの親を示す
        if self.par[x] ==x:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]

    def union(self,x,y):#x,yを繋げる
        px = self.find(x)
        py = self.find(y)
        if px == py:
            return 
        if self.rank[px]<self.rank[py]:
            self.par[px]=py
        else:
            self.par[py]=px
        if self.rank[px]==self.rank[py]:
            self.rank[px] +=1
    
    def connect(self,x,y):#親が同じかみる
        return self.find(x)==self.find(y)
#クラスカル法
kei = Unionfindtree(N)
graph.sort()
dist =0
for d,i,j in graph:
    if not kei.connect(i,j):
        kei.union(i,j)
        dist +=d
print(dist)