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.PriorityQueue;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn=new Scanner(System.in);
		int n=scn.nextInt();
		int m=scn.nextInt();
		char [][]arr =new char[n][m];
		int sr =scn.nextInt();
		int sc =scn.nextInt();
		sr--;
		sc--;
		int dr =scn.nextInt();
		int dc=scn.nextInt();
		dr--;
		dc--;
		int cnt=0;
		while(cnt<n){
			String s =scn.next();
			arr[cnt]=s.toCharArray();
			cnt++;
		}
		System.out.println(find(sr,sc,arr,dr,dc));

	}
	// I have 5 choices either go any direction and if any one is block use the power...
	public static long find(int sr ,int sc , char[][]arr ,int dr ,int dc){
		PriorityQueue<pair>pq =new PriorityQueue<>();
		pq.add(new pair(sr,sc,0));
		int [][]arr1 =new int[arr.length][arr[0].length];
		int []r={0,0,1,-1};
		int []c={1,-1,0,0};
		int [][]dir={{-2,-2},{-2,-1},{-2,0},{-2,1},{-2,2},
					{-1,-2},{-1,-1},{-1,0},{-1,1},{-1,2},
					 {0,-2},{0,-1},{0,0},{0,1},{0,2},
					 {1,-2},{1,-1},{1,0},{1,1},{1,2},
					 {2,-2},{2,-1},{2,0},{2,1},{2,2}
					 };
		while(pq.size()>0){
			pair p1 =pq.remove();
			if(p1.r==dr&& p1.c==dc){
				return p1.cost;
			}
			if(arr1[p1.r][p1.c]==1){
				continue;
			}
			arr1[p1.r][p1.c]=1;
			// I have 4 directions choice and a magic choice
			for(int i=0;i<4;i++){
				int nr =p1.r+r[i];
				int nc=p1.c+c[i];
				if(isvalid(nr,nc,arr,arr1)){
					pq.add(new pair(nr,nc,p1.cost));
				}
			  
			}
			// using the power boom......
			for(int i=0;i<dir.length;i++){
			  int nr= dir[i][0]+p1.r;
			  int nc=dir[i][1]+p1.c;
			  if(isvalid(nr,nc,arr,arr1)){
					pq.add(new pair(nr,nc,p1.cost+1));
				}
			}
			 
		}
		
		return -1;
	}
	public static boolean isvalid(int r ,int c, char [][]arr ,int [][]arr1){
		if(r<0 || r>=arr.length ||c>=arr[0].length || c<0 || arr[r][c]=='#' ||arr1[r][c]==1 ){
			return false;
		}
		return true;
	}
	
	public static class pair implements Comparable<pair>{
		int r;
		int c;
		 long cost;
		 public pair(int r ,int c , long cost){
			 this.r=r;
			 this.c=c;
			 this.cost=cost;
			 
		 }
		 public int compareTo(pair o){
			 return (int)(this.cost - o.cost);
		 }
	}

}