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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

import sys
from collections import deque
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0

H,W = map(int,readline().split())
sh,sw = map(int,readline().split())
gh,gw = map(int,readline().split())

maze = []
for _ in range(H):
    res = [-1]*W
    s = readline().rstrip().decode('utf-8')
    for i in range(W):
        if s[i] == ".":
            res[i] = 0
    maze.append(res)
    #maze += [list(readline().rstrip().decode().replace(".",0).replace("#",-1))]
    

ct = 1
for h in range(H):
    for w in range(W):
        if maze[h][w] == 0:
            y,x = h,w
            maze[y][x] = ct
            stack = deque([[y,x]])
            while stack:
                i,j = stack.pop()
                for ni,nj in [(0,1),(1,0),(0,-1),(-1,0)]:
                    if not 0 <= i + ni < H or not 0 <= j + nj < W:
                        continue
                    if maze[i+ni][j+nj] == 0:
                        maze[i+ni][j+nj] = ct
                        stack.append([i+ni,j+nj])
            ct += 1

g = [[] for i in range(ct)]
g2 = [[0]*ct for _ in range(ct)]
for i in range(ct):
    g2[i][i] = 1
for h in range(H):
        for w in range(W):
            if maze[h][w] != -1:
                for i in range(3):
                    for j in range(-2,3):
                        if not 0 <= i + h < H or not 0 <= j + w < W:
                            continue
                        if maze[h+i][w+j] != -1 and \
                            not g2[maze[h][w]][maze[h+i][w+j]]:
                            g[maze[h][w]].append((1,maze[h+i][w+j]))
                            g[maze[h+i][w+j]].append((1,maze[h][w]))
                            g2[maze[h][w]][maze[h+i][w+j]] = 1
                            g2[maze[h+i][w+j]][maze[h][w]] = 1


#ダイクストラ法通常版
#O(ElogV)
import heapq
def dijkstra_heap(s,edge): #s:始点,g:グラフ
    #始点sから各頂点への最短距離
    d = [float("inf")] * len(edge)
    used = [True] * len(edge) #True:未確定
    d[s] = 0
    used[s] = False
    edgelist = []
    for e in edge[s]:
        heapq.heappush(edgelist,e)
    while len(edgelist):
        minedge = heapq.heappop(edgelist)
        #まだ使われてない頂点の中から最小の距離のものを探す
        if not used[minedge[1]]: #既に最小距離が確定しているなら
            continue
        v = minedge[1] #vに頂点を代入
        d[v] = minedge[0] #その頂点のコストに重みを代入
        used[v] = False #vには最小距離が入っていることが証明できる(ヒープキューを使用しているため)
        for e in edge[v]: #頂点vから伸びている辺の先をヒープキューに入れてく
            if used[e[1]]:
                heapq.heappush(edgelist,(e[0]+d[v],e[1]))
    return d #回しきったらreturn(たどり着けなかった頂点はinf.)


start = maze[sh-1][sw-1]
goal = maze[gh-1][gw-1]

res = dijkstra_heap(start,g)

if res[goal] == float("inf"):
    print(-1)
else:
    print(res[goal])