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)

#!/usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
import itertools
sys.setrecursionlimit(10**5)
stdin = sys.stdin
bisect_left = bisect.bisect_left
bisect_right = bisect.bisect_right
def LI(): return list(map(int, stdin.readline().split()))
def LF(): return list(map(float, stdin.readline().split()))
def LI_(): return list(map(lambda x: int(x)-1, stdin.readline().split()))
def II(): return int(stdin.readline())
def IF(): return float(stdin.readline())
def LS(): return list(map(list, stdin.readline().split()))
def S(): return list(stdin.readline().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')

#A
def A():
    x, a, b = LI()
    if b <= a:
        print("delicious")
    elif b <= a + x:
        print("safe")
    else:
        print("dangerous")

    return

#B
def B():
    n = II()
    a = IR(n)
    check = [False] * (n + 1)
    check[1] = True
    now = a[0]
    for i in range(1,n+1):
        if check[now]:
            break
        if now == 2:
            print(i)
            return
        check[now] = True
        now = a[now - 1]
    print(-1)
    return

#C
def C():
    return

#D
def D():
    def find(x):
        if x == par[x]:
            return x
        par[x] = find(par[x])
        return par[x]
    def unite(x, y):
        x = find(x)
        y = find(y)
        if x == y: return False
        par[y] = x
        return True
    def solve(a, k):
        for i in range(n-1):
            q0, q1 = xy[i][k], xy[i + 1][k]
            z, v = xy[i][2], xy[i + 1][2]
            edg[a + i] = [q1 - q0, z, v]
    n = II()
    par = [i for i in range(n)]
    xy = [None] * n
    for i in range(n):
        x, y = LI()
        xy[i] = [x, y, i]
    xy.sort(key=lambda x: x[0])
    edg = [None] * (2 * n - 2)
    solve(0, 0)
    xy.sort(key=lambda x: x[1])
    solve(n - 1, 1)
    ans = 0
    a = 0
    edg.sort(key=lambda x: x[0])
    for value, p0, p1 in edg:
        if a == n - 1:
            break
        if unite(p0, p1):
            a += 1
            ans += value
    print(ans)
    return

#Solve
if __name__ == '__main__':
    D()