Problem:
Score : 400 points

Problem Statement

A maze is composed of a grid of H \times W squares - H vertical, W horizontal.

The square at the i-th row from the top and the j-th column from the left - (i,j) - is a wall if S_{ij} is # and a road if S_{ij} is ..

There is a magician in (C_h,C_w). He can do the following two kinds of moves:

Move A: Walk to a road square that is vertically or horizontally adjacent to the square he is currently in.

Move B: Use magic to warp himself to a road square in the 5\times 5 area centered at the square he is currently in.

In either case, he cannot go out of the maze.

At least how many times does he need to use the magic to reach (D_h, D_w)?

Constraints

1 \leq H,W \leq 10^3

1 \leq C_h,D_h \leq H

1 \leq C_w,D_w \leq W

S_{ij} is # or ..

S_{C_h C_w} and S_{D_h D_w} are ..

(C_h,C_w) \neq (D_h,D_w)

Input

Input is given from Standard Input in the following format:

H W
C_h C_w
D_h D_w
S_{11}\ldots S_{1W}
\vdots
S_{H1}\ldots S_{HW}

Output

Print the minimum number of times the magician needs to use the magic. If he cannot reach (D_h,D_w), print -1 instead.

Sample Input 1

4 4
1 1
4 4
..#.
..#.
.#..
.#..

Sample Output 1

1

For example, by walking to (2,2) and then using the magic to travel to (4,4), just one use of magic is enough.

Note that he cannot walk diagonally.

Sample Input 2

4 4
1 4
4 1
.##.
####
####
.##.

Sample Output 2

-1

He cannot move from there.

Sample Input 3

4 4
2 2
3 3
....
....
....
....

Sample Output 3

0

No use of magic is needed.

Sample Input 4

4 5
1 2
2 5
#.###
####.
#..##
#..##

Sample Output 4

2

Solution in Ruby (2.7.1) (judge verdict: Time Limit Exceeded)

require "set"

h, w = gets.split(" ").map(&:to_i)
ch, cw = gets.split(" ").map(&:to_i)
dh, dw = gets.split(" ").map(&:to_i)

# indexの関係で-1する
caxs = [ch-1, cw-1] # 現在地
daxs = [dh-1, dw-1] # 目的地

arr = []
maze = h.times.map {
  arr << [] # 初期化処理をついでにやる
  gets.chomp
}

queue = [caxs]
arr[ch-1][cw-1] = 0
(h*w).times do |i|
  queue2 = Set.new
  queue3 = Set.new
  # 上下左右の移動
  while true
    x, y = queue.pop

    # ゴールなら終了
    if [x, y] == daxs
      puts i
      return
    end
    # 上下左右のマスを確認する
    [[-1, 0], [1, 0], [0, -1], [0, 1]].each do |dx, dy|
      nx = x + dx
      ny = y + dy
      # 座標範囲外または既に探査済みの場合スキップする
      next unless 0 <= nx && nx < h && 0 <= ny && ny < w && !arr[nx][ny]

      if maze[nx][ny] == "."
        arr[nx][ny] = i
        queue << [nx, ny]
      else
        arr[nx][ny] = -1
        queue2 << [nx, ny]
      end
    end

    # 斜め方向の2マス先のワープを検証するために斜めに壁があるか確認する
    [[-1, -1], [1, 1], [1, -1], [-1, 1]].each do |dx, dy|
      nx = x + dx
      ny = y + dy
      # 座標範囲外の場合スキップする
      next unless 0 <= nx && nx < h && 0 <= ny && ny < w
      
      # いったんすべて格納する
      queue3 << [nx, ny]
      arr[nx][ny] = -1 if maze[nx][ny] == "#"
    end

    break if queue.length == 0
  end
  
  queue3.each do |x, y|
    queue2 << [x, y] if arr[x][y].nil? || arr[x][y] == -1
  end
  break if queue.length == 0 && queue2.length == 0

  tmp_queue = queue2.to_a
  # ワープ
  while true
    x, y = tmp_queue.pop

    [[-1, 0], [1, 0], [0, -1], [0, 1], [-1, 1], [1, -1], [-1, -1], [1, 1]].each do |dx, dy|
      nx = x + dx
      ny = y + dy
      # 座標範囲外または既に探査済みの場合スキップする
      next unless 0 <= nx && nx < h && 0 <= ny && ny < w && !arr[nx][ny]

      if maze[nx][ny] == "."
        arr[nx][ny] = i+1
        queue << [nx, ny]
      else
        arr[nx][ny] = -2
      end
    end

    break if tmp_queue.length == 0
  end
  
  break if queue.length == 0
end
# arr.each {|a| p a }
p -1