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)

import copy
import math
import time
import statistics
import math
import itertools
import bisect
import sys
from decimal import *
from collections import deque


def get_int():
    return int(input())

def get_string():
    return input()

def get_int_list():
    return [int(x) for x in input().split()]

def get_string_list():
    return input().split()

def get_int_multi():
    return map(int, input().split())

def get_string_char_list():
    return list(str(input()))

# print("{} {}".format(a, b))
# a_list = [0] * a
sys.setrecursionlimit(10 ** 6)

def main():
    start = time.time()

    h, w = get_int_list()
    sr, sc = get_int_list()
    er, ec = get_int_list()

    d = [[-1] * (w+2) for _ in range(h+2)]

    for i in range(h):
        wk = get_string()
        for ii in range(w):
            if wk[ii] == ".":
                d[i+1][ii+1] = 10 ** 6

    d[sr][sc] = 0
    q = deque([])
    q.append([sr,sc])

    def idou(i,ii,cnt_mahou):
        if d[i][ii] == 10 ** 6:
            q.append([i, ii])
            d[i][ii] = cnt_mahou

    ans = -1
    flg = True
    finish = False
    while flg:
        flg = False
        while q:
            wk = q.popleft()
            i = wk[0]
            ii = wk[1]
            cnt_mahou = d[i][ii]

            idou(i - 1, ii, cnt_mahou)
            idou(i + 1, ii, cnt_mahou)
            idou(i, ii - 1, cnt_mahou)
            idou(i, ii + 1, cnt_mahou)

            if d[er][ec] != 10 ** 6:
                ans = d[er][ec]
                finish = True
                break

        if finish:
            break

        for iii in range(h):
            for iiii in range(w):
                if cnt_mahou == d[iii][iiii]:

                    # 魔法を使う
                    for wk in range(-2, 3):
                        for wk2 in range(-2, 3):
                            if abs(wk) + abs(wk2) > 1:
                                flg = True
                                idou(iii + wk, iiii + wk2, cnt_mahou + 1)




    print(ans)

    print(time.time() - start)


if __name__ == '__main__':
    main()