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: Time Limit Exceeded)

from collections import deque
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(input()))

# コストをまとめる表
cost = [[float("inf")] * W for i in range(H)]

def bfs(que, c):
    # 幅優先探索を用いて、位置(init_x, init_y)から追加コストなしで辿り着けるマスを列挙

    next_que = deque([])

    while que:

        # キューの先頭を取り出す
        x, y = que.popleft()
        cost[x][y] = c

        # 取り出してきたキューがゴールなら探索をやめる
        if (x, y) == (D_h-1, D_w-1):
            return True, next_que

        moveA_vector = [(1,0),(-1,0),(0,1),(0,-1)]

        # 現在地から移動Aで到達可能で未探索の点をキューに追加
        # 移動4方向をループ
        for dx, dy in moveA_vector:
            nx = x + dx
            ny = y + dy
            if 0 <= nx < H and 0 <= ny < W and maze[nx][ny] != "#" and cost[nx][ny] == float("inf"):
                que.append((nx, ny))
                # cost[nx][ny] = c この処理は先頭でやる

        # 現在地から移動Bで到達可能で未探索の点を次のコストのキューに追加
        for dx in range(-2, 3):
            for dy in range(-2, 3):
                if (dx, dy) not in moveA_vector:  # 5*5の範囲内でかつ移動Aで到達できない点 = 現在地から移動Bで到達可能な点
                    nx = x + dx
                    ny = y + dy
                    if 0 <= nx < H and 0 <= ny < W and maze[nx][ny] != "#" and cost[nx][ny] == float("inf"):  # 到達可能で未探索の点
                        if ((nx, ny) not in next_que) and ((nx+1, ny) not in next_que) and ((nx-1, ny) not in next_que) and ((nx, ny+1) not in next_que) and ((nx, ny-1) not in next_que):
                            # next_queにすでに入っている点でなく、next_queにすでに入っている点から移動Aで到達できる点でもない
                            next_que.append((nx, ny))

    return False, next_que

c = 0
que = deque([])
que.append((C_h-1, C_w-1))
while True:
    f, que = bfs(que, c)
    if f:  # ゴールが見つかった
        print(c)
        sys.exit()
    if not que:  # queが空のとき、これ以上移動不可能（dequeなのでque==[]は誤り）
        break
    c += 1
print("-1")