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 Python (3.4.3) (judge verdict: Accepted)

#from numpy import* #import numpy as np
#from scipy.sparse.csgraph import shortest_path #shortest_path(csgraph=graph) #dijkstra #floyd_warshall
#from scipy.sparse import csr_matrix

from collections import* #defaultdict Counter deque appendleft
from fractions import gcd
from functools import* #reduce
from itertools import* #permutations("AB",repeat=2) combinations("AB",2) product("AB",2) groupby accumulate
from operator import mul,itemgetter
from bisect import* #bisect_left bisect_right
from heapq import* #heapify heappop heappushpop
from math import factorial,pi
from copy import deepcopy
import sys
sys.setrecursionlimit(10**8)
#input=sys.stdin.readline  #危険！基本オフにしろ！

class UNION_FIND(object):
    def __init__(self,n):
        self.parent=[-1 for i in range(n)]
        
    def root(self,x):
        if self.parent[x]<0:
            return x
        else:
            self.parent[x]=self.root(self.parent[x])
            return self.parent[x]
 
    def size(self,x):
        return -self.parent[self.root(x)]
    def union(self,x,y):
        x=self.root(x)
        y=self.root(y)
        if x==y:
            return False
        if self.size(x)<self.size(y):
            x,y=y,x
        self.parent[x]+=self.parent[y]
        self.parent[y]=x
        return True

def KRUSKAL(n, edges, TF_of_maximize):
    edges.sort(key=lambda x: x[2],reverse=TF_of_maximize)
    uf = UNION_FIND(n)
    cost = []
    for (a,b,c) in edges:
        if uf.root(a)!= uf.root(b):
            uf.union(a,b)
            cost.append(c)
    return cost


def main():
    n=int(input())
    inp=[[i]+list(map(int,input().split()))for i in range(n)]
    inpX=sorted(inp,key=lambda x:x[1])
    inpY=sorted(inp,key=lambda x:x[2])
    edges=[]
    for i in range(1,n):
        a,b,c=inpX[i]
        d,e,f=inpX[i-1]
        edges.append((a,d,min(abs(b-e),abs(c-f))))
        a,b,c=inpY[i]
        d,e,f=inpY[i-1]
        edges.append((a,d,min(abs(b-e),abs(c-f))))
    
    print(sum(KRUSKAL(n, edges, 0))) #頂点数 辺(コスト) #最大化するか否か 
    

if __name__ == '__main__':
    main()