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.ArrayDeque;
import java.util.HashSet;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;

public class Main {
    
    private static class Point {
        private final int i;
        private final int j;
        
        private Point(int i, int j) {
            this.i = i;
            this.j = j;
        }
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        final int H = sc.nextInt();
        final int W = sc.nextInt();
        final int startI = sc.nextInt() - 1;
        final int startJ = sc.nextInt() - 1;
        final int endI = sc.nextInt() - 1;
        final int endJ = sc.nextInt() - 1;
        sc.nextLine();
        
        final int[][] field = new int[H][W];
        
        for (int i = 0; i < H; i++) {
            String s = sc.nextLine();
            
            for (int j = 0; j < W; j++) {
                field[i][j] = s.charAt(j) == '.'
                        ? -1 // 未検証
                        : -2; //壁
            }
        }
        
        int currJump = 0;
        Queue<Point> testTarget = new ArrayDeque<>();
        testTarget.add(new Point(startI, startJ));
        
        while (true) {
            Set<Point> pointsOfCurrJump = new HashSet<>();
            
            // ジャンプ無しで到達できる位置をすべて調べ上げる
            while (!testTarget.isEmpty()) {
                Point p = testTarget.poll();
                pointsOfCurrJump.add(p);
                
                if (1 <= p.i && field[p.i - 1][p.j] == -1) {
                    field[p.i - 1][p.j] = currJump;
                    testTarget.add(new Point(p.i - 1, p.j));
                }
                if (1 <= p.j && field[p.i][p.j - 1] == -1) {
                    field[p.i][p.j - 1] = currJump;
                    testTarget.add(new Point(p.i, p.j - 1));
                }
                if (p.i <= H - 2 && field[p.i + 1][p.j] == -1) {
                    field[p.i + 1][p.j] = currJump;
                    testTarget.add(new Point(p.i + 1, p.j));
                }
                if (p.j <= W - 2 && field[p.i][p.j + 1] == -1) {
                    field[p.i][p.j + 1] = currJump;
                    testTarget.add(new Point(p.i, p.j + 1));
                }
            }
            
            // ジャンプ無しで到達できる場所を調べつくしたら、
            // そこからジャンプして到達できる位置をすべて調べ上げる
            int j = ++currJump;
            pointsOfCurrJump.forEach(p -> {
                for (int di = -2; di <= 2; di++) {
                    for (int dj = -2; dj <= 2; dj++) {
                        int ni = p.i + di;
                        int nj = p.j + dj;
                        
                        if (0 <= ni && ni < H && 0 <= nj && nj < W && field[ni][nj] == -1) {
                            field[ni][nj] = j;
                            testTarget.add(new Point(ni, nj));
                        }
                    }
                }
            });
            if (field[endI][endJ] != -1 || testTarget.isEmpty()) {
                break;
            }
        }
        System.out.println(field[endI][endJ]);
    }
}