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 PHP (7.4.4) (judge verdict: Accepted)

<?php
list($n) = ints();
for ($i = 0; $i < $n; ++$i) {
    list($x[], $y[]) = ints();
    $xi[] = $i;
    $yi[] = $i;
}
array_multisort($x, SORT_ASC, $xi);
array_multisort($y, SORT_ASC, $yi);
$kr = new Kruskal($n);
for ($i = 0; $i < $n - 1; ++$i) {
    $kr->connect($xi[$i], $xi[$i + 1], abs($x[$i + 1] - $x[$i]));
    $kr->connect($yi[$i], $yi[$i + 1], abs($y[$i + 1] - $y[$i]));
}
echo $kr->solve();
function ints()
{
    return array_map('intval', explode(' ', trim(fgets(STDIN))));
}
class Kruskal
{
    private $X, $Y, $W;
    private $uft;

    function __construct($n)
    {
        $this->uft = new UnionFind($n);
    }

    function connect($x, $y, $w)
    {
        $this->X[] = $x;
        $this->Y[] = $y;
        $this->W[] = $w;
    }

    function solve()
    {
        $costSum = 0;
        array_multisort($this->W, $this->X, $this->Y);
        foreach ($this->W as $k => $w) {
            if (!$this->uft->isSame($this->X[$k], $this->Y[$k])) {
                $this->uft->unite($this->X[$k], $this->Y[$k]);
                $costSum += $w;
            }
        }
        return $costSum;
    }
}

class UnionFind
{
    private $par; // par[$i]:$iの親 自分が根の場合は自身のindexとなる
    private $size;

    function __construct($n)
    {
        for ($i = 0; $i < $n; $i++) {
            $this->par[$i] = $i; // 最初は全てが根
            $this->size[$i] = 1;
        }
    }

    function root($x)
    { // 根を返す
        if ($this->par[$x] === $x) return $x; // 根の場合
        return $this->par[$x] = $this->root($this->par[$x]); // 全ての枝の親を根にしながら再帰処理
    }

    function unite($x, $y)
    { // $xと$yの木を併合
        $ix = $this->root($x);
        $iy = $this->root($y);
        if ($ix === $iy) return;
        if ($this->size[$ix] < $this->size[$iy]) list($ix, $iy) = [$iy, $ix];
        $this->size[$ix] += $this->size[$iy];
        $this->par[$iy] = $ix; // $yの根を$xの根に付ける
    }

    function isSame($x, $y)
    { // $xと$yの根が同じか
        return $this->root($x) === $this->root($y);
    }

    function size($x)
    {
        return $this->size[$this->root($x)];
    }
}