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)

use std::io::*;
use std::str::FromStr;
use std::collections::VecDeque;

//https://qiita.com/tubo28/items/e6076e9040da57368845
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

//移動方向のベクトルを配列で用意する
const DI:[isize;4] = [-1,0,1,0];
const DJ:[isize;4] = [0,-1,0,1];

fn main() {
    let h: usize = read();
    let w: usize = read();
    let C:(usize,usize) = (read::<usize>()-1,read::<usize>()-1);
    let D:(usize,usize) = (read::<usize>()-1,read::<usize>()-1);
    let map: Vec<Vec<char>> = (0..h).map(|_| read::<String>().chars().collect()).collect();

    /*
    println!("{:?}",C);
    println!("{:?}",D);
    println!("{:?}",map);
    */

    //h行w列のベクタで、i64の最大値で初期化
    let mut dist:Vec<Vec<i64>> = vec![vec![std::i64::MAX; w]; h];
    //0-1BFS用のdequeを用意する
    let mut q: VecDeque<(usize,usize)> = VecDeque::new();
    //スタート地点のコストを0に設定
    dist[C.0][C.1] = 0;
    q.push_front((C.0,C.1));

    while !q.is_empty(){
        //dequeの先頭要素を取り出す
        let i = q.front().unwrap().0;
        let j = q.front().unwrap().1;
        q.pop_front();
        //探索点のコストを設定
        let d = dist[i][j];
        //四方への移動を試す（移動A）
        for v in 0..4{
            let ni = i as isize + DI[v];
            let nj = j as isize + DJ[v];
            //移動先がmap内に収まるか判定
            if ni < 0 || ni >= h as isize || nj < 0 || nj >= w as isize{
                continue;
            }
            let ni = ni as usize;
            let nj = nj as usize;
            //移動先が壁でないか判定
            if map[ni][nj] == '#' {
                continue;
            }
            //移動先のコストが現在コストより小さいか判定
            if dist[ni][nj] <= d{
                continue;
            }
            //最小コストを更新
            dist[ni][nj] = d;
            //コスト0側に座標を追加
            q.push_front((ni,nj));

        }
        //ワープを試す（移動B）
        //近傍2x2の周りへワープ可能なので、for2重ループで確認する
        for ei in -2..=2{
            for ej in -2..=2{
                let ni = i as isize + ei;
                let nj = j as isize + ej;
                //移動先がmap内に収まるか判定
                if ni < 0 || ni >= h as isize || nj < 0 || nj >= w as isize{
                    continue;
                }
                let ni = ni as usize;
                let nj = nj as usize;
                //移動先が壁でないか判定
                if map[ni][nj] == '#' {
                    continue;
                }
                //移動先のコストが現在コスト+1(ワープの移動コスト追加)より小さいか判定
                if dist[ni][nj] <= d+1{
                    continue;
                }
                //最小コストを更新
                dist[ni][nj] = d+1;
                //コスト1側に座標を追加
                q.push_back((ni,nj));
            }
        }
    }
    let ans = dist[D.0][D.1];
    if ans == std::i64::MAX {
        println!("-1");
    }else{
        println!("{}",ans);
    }


}