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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(non_snake_case)]
use proconio::input;
// use std::collections::HashMap;
use std::collections::VecDeque;

fn main() {
    input! {
        H: usize,
        W: usize,
        Cs: [usize; 2],
        Ds: [usize; 2],
        Ss: [String; H],
    }

    let start = ((Cs[0] - 1) as isize, (Cs[1] - 1) as isize);
    let goal = ((Ds[0] - 1) as isize, (Ds[1] - 1) as isize);

    // 必要なワープ回数の最小値
    let mut warp_cnt: Vec<Vec<i64>> = Vec::new();
    for _h in 0..H {
        let arr = vec![std::i64::MAX; W];
        warp_cnt.push(arr);
    }

    let mut queue: VecDeque<(isize, isize)> = VecDeque::new();
    queue.push_back(start);
    warp_cnt[start.0 as usize][start.1 as usize] = 0;

    let dhws: Vec<(isize, isize)> = vec![(0, 1), (0, -1), (-1, 0), (1, 0)];
    let ds_magic: Vec<isize> = vec![-2, -1, 0, 1, 2];

    // WAで気付いたことメモ (21:56)
    // まず、歩き → 魔法 → 歩き →の順にFIFOでキューに入れると、まず魔法で行った後歩いて更新することになって、LTE
    // かと言って、歩き歩き歩き魔法魔法…の順に入れていくと、最後が魔法になって歩きで行けるはずのに余分にワープしている…?

    while !queue.is_empty() {
        let node: (isize, isize) = queue.pop_front().unwrap();

        for &(dh, dw) in dhws.iter() {
            let next_h = node.0 + dh;
            let next_w = node.1 + dw;


            if 0 <= next_h && next_h < H as isize && 0 <= next_w && next_w < W as isize {
                let next_is_road = &Ss[next_h as usize][next_w as usize..(next_w + 1) as usize] ==
                    ".";
                if !next_is_road {
                    continue;
                }

                let new_cnt = warp_cnt[node.0 as usize][node.1 as usize];
                let old_cnt = warp_cnt[next_h as usize][next_w as usize];

                if new_cnt < old_cnt {
                    // このnodeに行く甲斐がある
                    warp_cnt[next_h as usize][next_w as usize] = new_cnt;

                    // 魔法より歩き優先
                    queue.push_front((next_h, next_w));
                }
            }
        }

        // 魔法で行く場合
        for dh in ds_magic.iter() {
            for dw in ds_magic.iter() {

                let next_h = node.0 + dh;
                let next_w = node.1 + dw;


                if 0 <= next_h && next_h < H as isize && 0 <= next_w && next_w < W as isize {
                    let next_is_road =
                        &Ss[next_h as usize][next_w as usize..(next_w + 1) as usize] == ".";
                    if !next_is_road {
                        continue;
                    }

                    let new_cnt = warp_cnt[node.0 as usize][node.1 as usize] + 1;
                    let old_cnt = warp_cnt[next_h as usize][next_w as usize];

                    if new_cnt < old_cnt {
                        // このnodeに行く甲斐がある
                        warp_cnt[next_h as usize][next_w as usize] = new_cnt;
                        queue.push_back((next_h, next_w));
                    }
                }
            }
        }
    }

    let ans_cand = warp_cnt[goal.0 as usize][goal.1 as usize];
    let ans = if ans_cand == std::i64::MAX {
        -1
    } else {
        ans_cand
    };
    println!("{}", ans);
}