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 Python (3.8.2) (judge verdict: Wrong Answer)

# https://yamakasa.net/atcoder-abc-176-d/
# 迷路の入力をsys.stdinを使ってやる

from collections import deque
from sys import stdin
import sys

# 入力
H, W = map(int, input().split())
C_h, C_w = map(int, input().split())
D_h, D_w = map(int, input().split())
maze = []
for i in range(H):
    maze.append(list(stdin.readline()[:-1]))

# IDをまとめる表、-1で初期化
ID = [[-1] * W for i in range(H)]

# 移動Aで動ける範囲のマスは同一のIDを持つとして、IDを割り振る
id = 0
for x in range(H):
    for y in range(W):
        if maze[x][y]  == "." and ID[x][y] == -1:  # 到達可能でまだIDが割り振られていない点
            #ID[x][y] = id
            # BFSで同一のIDを持つ領域をくまなく探索
            que = deque([])
            que.append((x, y))
            while que:
                temp_x, temp_y = que.popleft()
                ID[temp_x][temp_y] = id
                for dx, dy in [(1,0), (-1,0), (0,1), (0,-1)]:
                    nx = temp_x + dx
                    ny = temp_y + dy
                    if 0 <= nx < H and 0 <= ny < W and maze[nx][ny] == "." and ID[nx][ny] == -1:
                        que.append((nx, ny))

            id += 1

# ワープ可能なID同士の繋がりを記録
ID_graph = [[0] * id for i in range(id)]  # IDmとIDnの間がワープ可能ならID[m][n] = 1、そうでなければID[m][n] = 0
for x in range(H):
    for y in range(W):
        if maze[x][y] != "#" and ID[x][y] != -1:  # IDが割り振られている点
            for dx in range(-2, 3):
                for dy in range(-2, 3):
                    if abs(dx) + abs(dy) >= 2:  # 移動Aでは移動不可で移動Bでは移動可能なマス
                        nx = x + dx
                        ny = x + dy
                        if 0 <= nx < H and 0 <= ny < W and maze[nx][ny] != "#" and ID[x][y] != ID[nx][ny]:
                            ID_graph[ID[x][y]][ID[nx][ny]] = 1
                            ID_graph[ID[nx][ny]][ID[x][y]] = 1

# ID間の関係から最短距離を求める
ID_score = [float("inf")] * id  # 各IDまで何回のワープでたどり着けるか、INFで初期化
init_ID = ID[C_h-1][C_w-1]  # 初期位置
ID_score[init_ID] = 0

que = deque([])
que.append(init_ID)
while que:
    temp_id = que.popleft()
    if temp_id == ID[D_h-1][D_w-1]:  # ゴールのIDに到達
        break
    for k in range(id):
        if ID_graph[temp_id][k] == 1 and ID_score[k] == float("inf"):  # 今着目しているIDからワープできてまだ見ていないID
            que.append(k)
            ID_score[k] = ID_score[temp_id] + 1

if ID_score[ID[D_h-1][D_w-1]] != float("inf"):
    print(ID_score[ID[D_h-1][D_w-1]])
else:
    print(-1)