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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.concurrent.PriorityBlockingQueue;

/**
 * @author b
 *
 */
public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner();
		PrintWriter writer = new PrintWriter(System.out);
		solve(scanner, writer);
		writer.close();
	}

	private static void solve(Scanner scanner, PrintWriter writer) {

		int rows = scanner.nextInt(), cols = scanner.nextInt();
		int startRow = scanner.nextInt() - 1, startCol = scanner.nextInt() - 1;
		int endRow = scanner.nextInt() - 1, endCol = scanner.nextInt() - 1;

		boolean[][] grid = new boolean[rows][cols];
		for (int i = 0; i < rows; i++) {
			char[] line = scanner.next().toCharArray();
			for (int j = 0; j < cols; j++) {
				grid[i][j] = line[j] == '.';
			}
		}

		// -1 havent been there
		// 0 can reach without magic
		// x requires x amount of magic
		int[][] magicCounts = new int[rows][cols];
		for (int i = 0; i < rows; i++)
			Arrays.fill(magicCounts[i], -1);

		magicCounts[startRow][startCol] = 0;
		Queue<Pair> toCheck = new PriorityBlockingQueue<>();
		toCheck.add(new Pair(startRow, startCol, 0));

		while (!toCheck.isEmpty()) {
			Pair next = toCheck.poll();
			int magicUsed = magicCounts[next.row][next.col];

			if (next.row == endRow && next.col == endCol) {
				break;
			}

			for (int i = -2; i <= 2; i++) {
				for (int j = -2; j <= 2; j++) {
					int r = next.row + i;
					int c = next.col + j;
					if (r < 0 || r >= rows)
						continue;
					if (c < 0 || c >= cols)
						continue;
					if (i == 0 && j == 0)
						continue;
					if (!grid[r][c])
						continue;

					int addMagic = 0;
					if (Math.abs(i) + Math.abs(j) != 1) {
						addMagic = 1;
					}

					if (magicCounts[r][c] == -1 || magicCounts[r][c] > magicUsed + addMagic) {
						magicCounts[r][c] = magicUsed + addMagic;
						toCheck.add(new Pair(r, c, magicCounts[r][c]));
					}
				}
			}
		}

		writer.println(magicCounts[endRow][endCol]);
	}

	static class Pair implements Comparable<Pair> {
		int row;
		int col;

		int priority;

		public Pair(int row, int col, int priority) {
			this.row = row;
			this.col = col;
			this.priority = priority;
		}

		@Override
		public int compareTo(Pair o) {

			return this.priority - o.priority;
		}
	}

	// Props to SecondThread
	static class Scanner {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer tokenizer = new StringTokenizer("");

		String next() {
			while (!tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return tokenizer.nextToken();
		}

		long[] nextLongArr(int num) {
			long[] vals = new long[num];
			for (int i = 0; i < num; i++)
				vals[i] = nextLong();
			return vals;
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		int[] nextIntArr(int numberOfElements) {
			int[] values = new int[numberOfElements];
			for (int i = 0; i < numberOfElements; i++)
				values[i] = nextInt();
			return values;
		}

		long nextLong() {
			return Long.parseLong(next());
		}
	}

}