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 heapq import heappop, heapify
import sys
input = sys.stdin.buffer.readline

class UnionFind :
    """
    Union Find木
    ---
    頂点のインデックスは0~

    Attributes
    ---
    parent : list<int>
        親のリスト．parent[頂点のインデックス] = 親のインデックス
    height : list<int>
        各木の高さ
    size : list<int>
        各木の頂点数
    component : int
        連結成分の数
    """

    def __init__(self, size) :
        """
        Parameters
        ---
        size : int
            頂点数
        """
        self.parent = list(range(size))
        self.height = [0] * size
        self.size = [1] * size
        self.component = size

    def root(self, index) :
        """
        親のインデックスの取得

        Parameters
        ---
        index : int
            取得する頂点のインデックス

        Returns
        ---
        rootIndex : int
            指定した頂点の根のインデックス
        """
        if self.parent[index] == index :  # 根の場合
            return index
        rootIndex = self.root(self.parent[index])  # 葉の場合親の根を取得
        self.parent[index] = rootIndex  # 親の付け直し
        return rootIndex

    def union(self, index1, index2) :  # 結合
        """
        木の結合

        Parameters
        ---
        index1 : int
        index2 : int
            結合する頂点のインデックス
        """
        root1 = self.root(index1)
        root2 = self.root(index2)

        if root1 == root2 :  # 連結されている場合
            return

        self.component -= 1  # 連結成分を減らす

        if self.height[root1] < self.height[root2] :
            self.parent[root1] = root2  # root2に結合
            self.size[root2] += self.size[root1]
        else :
            self.parent[root2] = root1  # root1に結合
            self.size[root1] += self.size[root2]
            if self.height[root1] == self.height[root2] :
                self.height[root1] += 1
        return

    def isSameRoot(self, index1, index2) :
        """
        同じ木に属するかを判定する

        Parameters
        ---
        index1 : int
        index2 : int

        Returns
        ---
        boolean
        """
        return self.root(index1) == self.root(index2)

    def sizeOfSameRoot(self, index) :
        """
        指定した頂点の属する木の大きさを取得する
        """
        return self.size[self.root(index)]

    def getComponent(self) :
        """
        連結成分数を取得する
        """
        return self.component

N = int(input())
X = []
Y = []

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

X.sort()
Y.sort()
edges = []

for (u, fr), (v, to) in zip(X, X[1:]):
    edges.append((abs(u - v), fr, to,))
for (u, fr), (v, to) in zip(Y, Y[1:]):
    edges.append((abs(u - v), fr, to,))

heapify(edges)
tree = UnionFind(N)
ans = 0
while edges:
    cost, fr, to = heappop(edges)
    if not tree.isSameRoot(fr, to):
        ans += cost
        tree.union(fr, to)

print(ans)