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.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;

public class Main {
	static final long MOD=1000000007;//998244353;
	public static void main(String[] args) {
		PrintWriter out = new PrintWriter(System.out);
		InputReader sc=new InputReader(System.in);
		int H=sc.nextInt();
		int W=sc.nextInt();
		int ch=sc.nextInt()-1;
		int cw=sc.nextInt()-1;
		int dh=sc.nextInt()-1;
		int dw=sc.nextInt()-1;
		char[][] cs=new char[H][W];
		for (int i = 0; i < cs.length; i++) {
			cs[i]=sc.next().toCharArray();
		}
		long[][] L = new long[H][W];
        for(int i=0;i<H;i++){
        	for (int j = 0; j < W; j++) {
        		L[i][j] = -1;
			}
        }
        int[][] visited = new int[H][W];
        L[ch][cw] = 0;
        PriorityQueue<Edge> Q = new PriorityQueue<Edge>();
        Q.add(new Edge(kumi(ch, cw, W), 0));
        while(!Q.isEmpty()){
            Edge C = Q.poll();
            int h=C.to/W;
            int w=C.to%W;
            if(visited[h][w]==0){
                L[h][w] = C.v;
                visited[h][w] = 1;
                int[] dx= {1,-1,0,0};
        		int[] dy= {0,0,1,-1};
        		for (int j2 = 0; j2 < dy.length; j2++) {
					if (w+dx[j2]>=0&&w+dx[j2]<W&&h+dy[j2]>=0&&h+dy[j2]<H) {
						if (cs[h+dy[j2]][w+dx[j2]]=='.') {
							Q.add(new Edge(kumi(h+dy[j2], w+dx[j2], W),L[h][w]+0));
						}
					}
				}
        		int[] d= {-2,-1,0,1,2};
        				for (int j2 = 0; j2 < d.length; j2++) {
        					for (int k = 0; k < d.length; k++) {
        						if (Math.abs(d[j2])+Math.abs(d[k])<=1) {
        							continue;
        						}
        						if (w+d[j2]>=0&&w+d[j2]<W&&h+d[k]>=0&&h+d[k]<H) {
        							if (cs[h+d[k]][w+d[j2]]=='.') {
        								Q.add(new Edge(kumi(h+d[k], w+d[j2], W),L[h][w]+1));
        							}
        						}
        					}
        		}
            }
        }
		System.out.println(L[dh][dw]);
 	}
	static int kumi(int i, int j, int W){
        return i*W+j;//W列の２次配列を一次元にする
    }
	static class Edge implements Comparable<Edge>{
		int to;
		long v;
		public Edge(int to,long v) {
			this.to=to;
			this.v=v;
		}
		@Override
    	public boolean equals(Object obj) {
    		if(obj instanceof Edge) {
    			Edge other = (Edge) obj;
    			return other.to==this.to && other.v==this.v;
    		}
    		return false;
    	}//同値の定義
    	@Override
    	public int hashCode() {
    		return Objects.hash(this.to,this.v);
    	}
    	@Override
		  public int compareTo( Edge p2 ){
			 if (this.v>p2.v) {
				return 1;
			}
			 else if (this.v<p2.v) {
				return -1;
			}
			 else {
				return 0;
			}
		  }
	} 
	static class InputReader { 
		private InputStream in;
		private byte[] buffer = new byte[1024];
		private int curbuf;
		private int lenbuf;
		public InputReader(InputStream in) {
			this.in = in;
			this.curbuf = this.lenbuf = 0;
		}
 
		public boolean hasNextByte() {
			if (curbuf >= lenbuf) {
				curbuf = 0;
				try {
					lenbuf = in.read(buffer);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (lenbuf <= 0)
					return false;
			}
			return true;
		}
 
		private int readByte() {
			if (hasNextByte())
				return buffer[curbuf++];
			else
				return -1;
		}
 
		private boolean isSpaceChar(int c) {
			return !(c >= 33 && c <= 126);
		}
 
		private void skip() {
			while (hasNextByte() && isSpaceChar(buffer[curbuf]))
				curbuf++;
		}
 
		public boolean hasNext() {
			skip();
			return hasNextByte();
		}
 
		public String next() {
			if (!hasNext())
				throw new NoSuchElementException();
			StringBuilder sb = new StringBuilder();
			int b = readByte();
			while (!isSpaceChar(b)) {
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}
 
		public int nextInt() {
			if (!hasNext())
				throw new NoSuchElementException();
			int c = readByte();
			while (isSpaceChar(c))
				c = readByte();
			boolean minus = false;
			if (c == '-') {
				minus = true;
				c = readByte();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res = res * 10 + c - '0';
				c = readByte();
			} while (!isSpaceChar(c));
			return (minus) ? -res : res;
		}
 
		public long nextLong() {
			if (!hasNext())
				throw new NoSuchElementException();
			int c = readByte();
			while (isSpaceChar(c))
				c = readByte();
			boolean minus = false;
			if (c == '-') {
				minus = true;
				c = readByte();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res = res * 10 + c - '0';
				c = readByte();
			} while (!isSpaceChar(c));
			return (minus) ? -res : res;
		}
 
		public double nextDouble() {
			return Double.parseDouble(next());
		}
 
		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
 
		public long[] nextLongArray(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}
 
		public char[][] nextCharMap(int n, int m) {
			char[][] map = new char[n][m];
			for (int i = 0; i < n; i++)
				map[i] = next().toCharArray();
			return map;
		}
	}
}