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)

import sys
sys.setrecursionlimit(10**7)
from collections import defaultdict

import collections
import itertools
import operator


class UnionFind:
    def __init__(self, elems=None):
        class KeyDict(dict):
            def __missing__(self, key):
                self[key] = key
                return key

        self.parent = KeyDict()
        self.rank = collections.defaultdict(int)

        if elems is not None:
            for elem in elems:
                _, _ = self.parent[elem], self.rank[elem]

    def find(self, x):
        if self.parent[x] == x:
            return x
        else:
            self.parent[x] = self.find(self.parent[x])
            return self.parent[x]

    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if self.rank[x] < self.rank[y]:
            self.parent[x] = y
        else:
            self.parent[y] = x
        if self.rank[x] == self.rank[y]:
            self.rank[x] += 1

    def are_same(self, x, y):
        return self.find(x) == self.find(y)

    def grouper(self):
        roots = [(x, self.find(x_par)) for x, x_par in self.parent.items()]
        root = operator.itemgetter(1)
        for _, group in itertools.groupby(sorted(roots, key=root), root):
            yield [x for x, _ in group]


'''
def init(n):
    return list(range(n+1))#
def find(lis, x):
    if lis[x] == x:
        return x
    else:
        lis[x] = find(lis, lis[x])
        return lis[x]

def sameset(lis, x, y):
    return find(lis, x) == find(lis, y)

def union(lis, x, y):
    x = find(lis, x)
    y = find(lis, y)
    if x != y:
        lis[x] = y
'''

N = int(input())
#union_find_tree = init(N)

town = []

for i in range(N):
    x, y = map(int, input().split())
    town.append((i, x, y))

x_sorted_town = sorted(town, key=lambda x: x[1])
y_sorted_town = sorted(town, key=lambda x: x[2])

graph = []

pre_tx = x_sorted_town[0]
pre_ty = y_sorted_town[0]
pre_nx = town[pre_tx[0]]
pre_ny = town[pre_ty[0]]
for tx, ty in zip(x_sorted_town[1:], y_sorted_town[1:]):
    nx = town[tx[0]]
    ny = town[ty[0]]
    wx = min(abs(tx[1] - pre_tx[1]), abs(nx[2] - pre_nx[2]))
    wy = min(abs(ty[2] - pre_ty[2]), abs(ny[1] - pre_ny[1]))
    graph.append((nx[0], pre_nx[0], wx))
    graph.append((ny[0], pre_ny[0], wy))
    pre_tx = tx
    pre_ty = ty
    pre_nx = nx
    pre_ny = ny

uf = UnionFind()

cost = 0
for (s,t,w) in sorted(graph, key=lambda x: x[2]):
    if not uf.are_same(s,t):
        uf.unite(s,t)
        cost += w
print(cost)