Problem: 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.

Examples

Input

4 4
1 1
4 4
..#.
..#.
.#..
.#..


Output

1


Input

4 4
1 4
4 1
.##.


.##.


Output

-1


Input

4 4
2 2
3 3
....
....
....
....


Output

0


Input

4 5
1 2
2 5
.###
.
..##
..##


Output

2

Solution (in JAVA):

Solution:
import java.util.Scanner;
import java.util.Queue;
import java.util.ArrayDeque;
 
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		//初期化と入力
		int h = sc.nextInt();
		int w = sc.nextInt();
		int[] c = new int[2];
		int[] d = new int[2];
		char[][] s = new char[h+4][w+4];
		c[0] = sc.nextInt()+2;
		c[1] = sc.nextInt()+2;
		d[0] = sc.nextInt()+2;
		d[1] = sc.nextInt()+2;
		for(int i = 0; i < h+4; i++) {
			for(int j = 0; j < w+4; j++) {
				s[i][j] = '0';
			}
		}
		for(int i = 0; i < h; i++) {
			String temp = sc.next();
			for(int j = 0; j < w; j++) {
				s[i+2][j+2] = temp.charAt(j);
			}
		}
		boolean[][]	 check = new boolean[h+4][w+4];
		for(int i = 0; i < h+4; i++) {
			for(int j = 0; j < w+4; j++) {
				check[i][j] = true;
			}
		}
		Queue<Integer> kohoy = new ArrayDeque<>();
		Queue<Integer> kohox = new ArrayDeque<>();
		Queue<Integer> searchy = new ArrayDeque<>();
		Queue<Integer> searchx = new ArrayDeque<>();
		kohoy.add(c[0]-1);
		kohox.add(c[1]-1);
		int ans = 0;
		//ここまで初期化と入力↑
		
		while(true) {
			//ワープ候補先を（１回目はスタート地点）を探索
			while(true) {
				//System.out.println(kohoy.peek()+" "+kohox.peek());
				if(kohoy.peek()==null) {
					break;
				}
				run(s,check,kohoy.poll(),kohox.poll(),searchy,searchx);
				//System.out.println(kohoy.peek()+" "+kohox.peek());
			}
			/*for(int i = 0; i < h + 4; i++) {
				for(int j = 0; j < w + 4; j++) {
					if(check[i][j]==true) {
						System.out.print('a');
					}else {
						System.out.print('b');
					}
				}System.out.println();
			}*/
			//System.out.println();
			//ゴールが探索済みなら終了
			if(!check[d[0]-1][d[1]-1]) {
				break;
			}
			//探索した場所（serachy,searchx)からワープ先候補を探してキュー(kohoy,kohox)に入れる
			while(true) {
				if(searchy.peek()==null) {
					break;
				}
				warp(s,check,kohoy,kohox,searchy,searchx);
			}
			//探索する地点とワープ先がなくなって、ゴールが未探索ならばゴール不可能で終了
			if(searchy.peek()==null && kohoy.peek()==null && check[d[0]-1][d[1]-1]) {
				ans = -1;
				break;
			}
 
			++ans;
		}
		System.out.println(ans);
		
	}
	//候補キューから探索開始地点を取り出して探索．探索した場所を探索済みキューに入れる
	public static void run(char[][] map,boolean[][] check,int y, int x,Queue<Integer> sy,Queue<Integer> sx) {
		check[y][x] =false;
		sy.add(y);
		sx.add(x);
		if(map[y+1][x] == '.' && check[y+1][x] == true) {
			run(map,check,y+1,x, sy,sx);
		}if(map[y-1][x] == '.' && check[y-1][x] == true) {
			run(map,check,y-1,x, sy,sx);
		}if(map[y][x+1] == '.' && check[y][x+1] == true) {
			run(map,check,y,x+1, sy,sx);
		}if(map[y][x-1] == '.' && check[y][x-1] == true) {
			run(map,check,y,x-1, sy,sx);
		}
	}
	//探索済みキューから取り出しワープできる場所を探して候補キューに入れる．
	public static void warp(char[][] map,boolean[][] check,Queue<Integer> kohoy,Queue<Integer> kohox,Queue<Integer> sy,Queue<Integer> sx) {
		int y = sy.poll();
		int x = sx.poll();
		for(int i = 0; i < 5; i++) {
			for(int j = 0; j < 5; j++) {
				if(map[y-2+i][x-2+j] == '.' && check[y-2+i][x-2+j] == true) {
					kohoy.add(y-2+i);
					kohox.add(x-2+j);
					//System.out.println("a"+(y-2+i)+" "+(x-2+j));
				}
			}
		}
	}
}