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.io.*;
import java.math.*;
import java.util.*;
public class Main {
	/*BFS over adjacent road sections
	 * Country/Territory 
	 * If a square could be achieved through 0 magic, a hard square will take at most 1
	 * magic (Therefore BFS on numbers 0-upper bound); Until all valid road cells possible 
	 * to hit from start have been visited*/
	static long[][] dp;
	static int R;
	static int C;
	static boolean[][] vis;
	static char[][] grid;
	static Stack<int[]> coords;
	static Stack<int[]> next; //Use queue sizing instead (only one queue would be needed then)
	static boolean[][] addvis;
	public static void main(String[] args) {  
		FastScanner I = new FastScanner(); //Input
		OutPut O = new OutPut(); //Output
		R = I.nextInt();
		C = I.nextInt();
		int r1 = I.nextInt()-1;
		int c1 = I.nextInt()-1;
		int r2 = I.nextInt()-1;
		int c2 = I.nextInt()-1;
		grid = new char[R][C];
		addvis = new boolean[R][C];
		int cur = 0;
		vis = new boolean[R][C];
		for (int i = 0; i < R; i++) grid[i] = I.next().toCharArray();
		dp = new long[R][C];
		for (int i = 0; i < R; i++) {
			for (int j = 0; j < C; j++) dp[i][j] = Integer.MAX_VALUE;
		}
		dp[r1][c1]=0;
		coords = new Stack<int[]>();
		next = new Stack<int[]>();
		int[] begin = {r1,c1};
		coords.push(begin);
		addvis[r1][c1]=true;
		while (true) {
			boolean added=false;
			while (!coords.isEmpty()) {
				int[] now = coords.pop();
				int r = now[0];
				int c = now[1];
				if (dp[r][c]==cur) {
					added=true;
					BFS(now,cur);
				}
			}
			while (!next.isEmpty()) {
				coords.push(next.pop());
				//O.pln(coords.peek()[0]+" "+coords.peek()[1]);
			}
			cur++;
			if (!added) break;
		}
		O.pln(dp[r2][c2]==Integer.MAX_VALUE?-1:dp[r2][c2]);
	}
	public static void BFS(int[] start, int cur) {
		vis[start[0]][start[1]]=true;
		Queue<int[]> q = new LinkedList<int[]>();
		q.add(start);
		while (!q.isEmpty()) {
			int size = q.size();
			for (int i = 0; i < size; i++) {
				int[] pos = q.poll();
				int r = pos[0];
				int c = pos[1];
				dp[r][c]=cur;
				if (r-1>=0&&grid[r-1][c]=='.'&&!vis[r-1][c]) {
					int[] add = {r-1,c};
					vis[r-1][c]=true;
					dp[r-1][c]=cur; //Can go there without magic
					q.add(add);
				}
				if (r+1<R&&grid[r+1][c]=='.'&&!vis[r+1][c]) {
					int[] add = {r+1,c};
					vis[r+1][c]=true;
					dp[r+1][c]=cur; //Can go there without magic
					q.add(add);
				}
				if (c-1>=0&&grid[r][c-1]=='.'&&!vis[r][c-1]) {
					int[] add = {r,c-1};
					vis[r][c-1]=true;
					dp[r][c-1]=cur; //Can go there without magic
					q.add(add);
				}if (c+1<C&&grid[r][c+1]=='.'&&!vis[r][c+1]) {
					int[] add = {r,c+1};
					vis[r][c+1]=true;
					dp[r][c+1]=cur; //Can go there without magic
					q.add(add);
				}
				for (int backr = max(r-2,0); backr<=min(r+2,R-1); backr++) {
					for (int backc = max(c-2,0); backc<=min(c+2,C-1); backc++) {
						if (!vis[backr][backc]&&!addvis[backr][backc]&&grid[backr][backc]=='.') {
							int[] add = {backr,backc};
							next.push(add);
							addvis[backr][backc]=true;
							dp[backr][backc]=min(dp[backr][backc],cur+1);
						}
						//Maybe magic is best in some cases
					}
				}
			}
		}
	}
	public static long min(long a, long b) {return Math.min(a,b);}
	public static long max(long a, long b) {return Math.max(a,b);}
	public static int min(int a, int b) {return Math.min(a,b);}
	public static int max(int a, int b) {return Math.max(a,b);}
	public static long abs(long x, long y) {return Math.abs(x-y);}
	public static long ceil(long num, long den) {long ans = num/den; if (num%den!=0) 
	ans++; return ans;}
	public static long GCD(long a, long b) {
		if (a==0||b==0) return max(a,b);
		return GCD(max(a,b),max(a,b)%min(a,b));
	}
	public static long FastExp(long base, long exp, long mod) {
		long ans=1;
		while (exp>0) {
			if (exp%2==1) ans*=base;
			exp/=2;
			base*=base;
			base%=mod;
			ans%=mod;
		}
		return ans;
	}
	public static long ModInv(long num,long mod) {return FastExp(num,mod-2,mod);}
	public static int pop(long x) { //Returns number of bits within a number
		int cnt = 0;
		while (x>0) {
			if (x%2==1) cnt++;
			x/=2;
		}
		return cnt;
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());};
		double nextDouble() {return Double.parseDouble(next());}
	}
	static class OutPut{
		PrintWriter w = new PrintWriter(System.out);
		void pln(double x) {w.println(x);w.flush();}
		void pln(boolean x) {w.println(x);w.flush();}
		void pln(int x) {w.println(x);w.flush();}
		void pln(long x) {w.println(x);w.flush();}
		void pln(String x) {w.println(x);w.flush();}
		void pln(char x) {w.println(x);w.flush();}
		void pln(StringBuilder x) {w.println(x);w.flush();}
		void pln(BigInteger x) {w.println(x);w.flush();}
		void p(int x) {w.print(x);w.flush();}
		void p(long x) {w.print(x);w.flush();}
		void p(String x) {w.print(x);w.flush();}
		void p(char x) {w.print(x);w.flush();}
		void p(StringBuilder x) {w.print(x);w.flush();}
		void p(BigInteger x) {w.print(x);w.flush();}
		void p(double x) {w.print(x);w.flush();}
		void p(boolean x) {w.print(x);w.flush();}
	}
}