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)

#----------[インポート]----------#
from collections import deque

def main():
    #----------[入力受取・初期設定]----------#
    H, W = map(int, input().split())
    C = tuple(map(int, input().split())) # 魔法の位置
    D = tuple(map(int, input().split())) # ゴール
    D_y = D[0] -1 # 0スタート
    D_x = D[1] -1 # 0スタート
    S = [list(input()) for _ in range(H)]
    visited = [[-1] * W for _ in range(H)]
    visited[C[0]-1][C[1]-1] = 0 # 初期値
    moves = [(1, 0), (0, 1), (-1, 0), (0, -1)] # 普通の移動
    main_q = deque()
    magic_q = deque() # マジック用のキュー
    main_q.append((C[0]-1, C[1]-1))
    magic_q.append((C[0]-1, C[1]-1))

    #----------[BFS]----------#
    while main_q:
        if visited[D_y][D_x] != -1:
            break

        y, x = main_q.pop()
        for move in moves:
            dy, dx = move
            moved_y = y + dy
            moved_x = x + dx

            if moved_y < 0 or H -1 < moved_y or moved_x < 0 or W -1 < moved_x:
                continue
            if S[moved_y][moved_x] == '#':
                continue
            if S[moved_y][moved_x] == '.' and visited[moved_y][moved_x] == -1:
                main_q.append((moved_y, moved_x))
                magic_q.append((moved_y, moved_x))
                visited[moved_y][moved_x] = visited[y][x]

        #----------[マジック用のBFS]----------#
        if not main_q: # main_qが空になった場合は探索済みからマジックを使う
            while magic_q:
                y, x = magic_q.pop()
                for dy in range(-2, 3):
                    for dx in range(-2, 3):
                        moved_y = y + dy
                        moved_x = x + dx

                        if moved_y < 0 or H -1 < moved_y or moved_x < 0 or W -1 < moved_x:
                            continue
                        if S[moved_y][moved_x] == '#':
                            continue
                        if S[moved_y][moved_x] == '.' and visited[moved_y][moved_x] == -1:
                            main_q.append((moved_y, moved_x))
                            visited[moved_y][moved_x] = visited[y][x] + 1

    answer = visited[D_y][D_x]
    print(answer)


if __name__ == "__main__":
    main()