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.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Queue;
import java.util.Scanner;

public class Main {

	static int[][] map;
	static int[][] directions8= {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
	static int[][] directions25= {{-2,-2},{-2,-1},{-2,0},{-2,1},{-2,2},{-1,-2},{-1,-1},{-1,1},{-1, 2}
										,{0,-2},{0,2},{1,-2},{1,-1},{1,1},{1,2},{2,-2},{2, -1},{2,0},{2,1},{2,2}};
	static int[][] directions4= {{-1,0},{1,0},{0,-1},{0,1}};
	static int ans;
	static int[] parent;//union-find用
	static int[] rank;//union-find用
	static int[][] warp;
	static Queue<int[]> q;
	static int H;
	static int W;
	static int Dh;
	static int Dw;
	static char[][] S;

	public static void main(String[] args) throws IOException {

		Scanner sc = new Scanner(System.in);
		H = sc.nextInt();
		 W = sc.nextInt();
		int Ch = sc.nextInt()-1;
		int Cw = sc.nextInt()-1;
		 Dh = sc.nextInt()-1;
		 Dw = sc.nextInt()-1;

		 S = new char[H][W];
		 warp = new int[H][W];

		for(int i = 0;i<H ; i++) {
			Arrays.fill(warp[i], Integer.MAX_VALUE);
		}


		for(int i = 0;i<H ; i++) {
			String str = sc.next();
			for(int j = 0;j<W ; j++) {
				S[i][j] = str.charAt(j);
			}
		}

		q = new ArrayDeque<int[]>();
		ans = Integer.MAX_VALUE;
		warp[Ch][Cw] = 0;
		int[] c = {Ch, Cw,0};
		q.add(c);

		while(!q.isEmpty()) {
			int[] now = q.poll();
			int nowH = now[0];
			int nowW = now[1];
			int warped = now[2];

			for(int i = 0;i<4 ; i++) {
				walk(nowH+directions4[i][0], nowW+directions4[i][1], warped);
			}
			for(int i = 0;i< 20; i++) {
				walk(nowH+directions25[i][0], nowW+directions25[i][1], warped+1);
			}
		}
		if(ans == Integer.MAX_VALUE) {
			ans = -1;
		}

		System.out.println(ans);
	}


	public static void walk(int nowH, int nowW, int warped) {
		if(nowH < 0 || nowH >= H || nowW <0 || nowW >= W || S[nowH][nowW] == '#' || warped >= warp[nowH][nowW]) {
			return;
		}
		if(nowH == Dh && nowW == Dw && ans > warped) {
			ans = warped;
		}
		int[] now = {nowH, nowW, warped};
		q.add(now);
		warp[nowH][nowW] = warped;
		for(int i = 0;i<4 ; i++) {
			walk(nowH+directions4[i][0], nowW+directions4[i][1], warped);
		}
	}
//	public static void warp(int nowH, int nowW, int warped) {
//
//
//
//
//	}

	//素数判定
	public static boolean isPrime(long num) {
		if (num < 2) return false;
		else if (num == 2) return true;
		else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

		double sqrtNum = Math.sqrt(num);
		for (int i = 3; i <= sqrtNum; i += 2)
		{
			if (num % i == 0)
			{
				// 素数ではない
				return false;
			}
		}

		// 素数である
		return true;
	}

	//BFS用に二つの配列を足し算する
	static int[] addArrayElms(int[] a, int[] b) {
		int[] c = new int[a.length];
		for(int i = 0; i < a.length; i++) {
			c[i] = a[i] + b[i];
		}
		return c;
	}

	//	//二分探索
	//k <= num となる最小の配列要素kのインデックスを返す
	static private int binarySearch(long num, long[] orderedArray){
		int lowerBorder = -1;
		int upperBorder = orderedArray.length;
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedArray[mid]<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//二分探索
	//k <= num となる最小のList要素kのインデックスを返す
	static private int binarySearch(long num, ArrayList<Long> orderedList){
		int lowerBorder = -1;
		int upperBorder = orderedList.size();
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedList.get(mid)<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//aとbの最小公倍数を求める
	public static int gcd(int a, int b) {
		return b == 0 ? a: gcd(b, a % b);
	}
	public static long gcd(long a, long b) {
		return b == 0 ? a: gcd(b, a % b);
	}

	//Union-Find用の配列を要素数nで初期化
	static void ufInit(int n){
		parent = new int[n];
		rank = new int[n];
	}

	//Union-Findの要素の最上位の親（グループ名）を返す
	static int ufFind(int a) {
		if(parent[a]==a) {
			return a;
		}else {
			return ufFind(parent[a]);
		}
	}

	//Union-Find木を連結する
	static void ufUnite(int a, int b) {
		a = parent[a];
		b = parent[b];

		if(a==b) {
			return;
		}
		if(rank[a] > rank[b]){
			parent[b]=a;
		}else {
			parent[a]=b;
			if(rank[a]==rank[b]) {
				rank[b]++;
			}
		}
	}

	static boolean ufSame(int a, int b) {
		return ufFind(a)==ufFind(b);
	}
}