Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 4
1 1
4 4
..#.
..#.
.#..
.#..

Sample Output 1

1

For example, by walking to (2,2) and then using the magic to travel to (4,4), just one use of magic is enough.

Note that he cannot walk diagonally.

Sample Input 2

4 4
1 4
4 1
.##.
####
####
.##.

Sample Output 2

-1

He cannot move from there.

Sample Input 3

4 4
2 2
3 3
....
....
....
....

Sample Output 3

0

No use of magic is needed.

Sample Input 4

4 5
1 2
2 5
#.###
####.
#..##
#..##

Sample Output 4

2

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define forn(i,a,b) for(int i =a;i<b;i++)
#define fi first
#define se second
#define fast ios_base::sync_with_stdio(false);
using namespace std;

//for debugging 
/*
g++ -D_GLIBCXX_ASSERTIONS -DDEBUG -ggdb3 -std=c++14 
*/
int recur_depth = 0;
#ifdef DEBUG
#define dbg(x) {++recur_depth; auto x_=x; --recur_depth; cerr<<string(recur_depth, '\t')<<"\e[91m"<<__func__<<":"<<__LINE__<<"\t"<<#x<<" = "<<x_<<"\e[39m"<<endl;}
#else
#define dbg(x)
#endif
template<typename Ostream, typename Cont>
typename enable_if<is_same<Ostream,ostream>::value, Ostream&>::type operator<<(Ostream& os,  const Cont& v){
	os<<"[";
	for(auto& x:v){os<<x<<", ";}
	return os<<"]";
}
template<typename Ostream, typename ...Ts>
Ostream& operator<<(Ostream& os,  const pair<Ts...>& p){
	return os<<"{"<<p.first<<", "<<p.second<<"}";
}
// debugging ends here

typedef long long int ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<ll> vl;
const int mxn = 1e3+99;
// const int inf = 1e9;
int cache[mxn][mxn];
int h,w;
char maze[mxn][mxn];
const int mxa = 1e6+10; 
set<int> adj[mxa];

int dx[4] = {-1,0,0,1},dy[4]={0,-1,1,0};
bool pos(int x, int y){
	if(x<0 || x>=h) return false;
	if(y<0 || y >=w) return false;
	if(maze[x][y]=='#') return false;
	return true;
}

void dfs(int x,int y,int com){
	// cout << x << ""  << y << endl;
	cache[x][y] = com;
	for(int i =0;i<4;i++){
		int nx = x+dx[i];
		int ny = y+dy[i];
		if(pos(nx,ny) && cache[nx][ny]==-1){
			cache[nx][ny] = com;
			dfs(nx,ny,com);
		}
	}
}
int ans =1e9;

vector<int>  vis;

void bfs(int u,int t,int len){
	vis[u] = 1;
	if(u==t){
		ans = min(ans,len);
		return;
	}

	for(auto v:adj[u]){
		if(!vis[v]){
			bfs(v,t,len+1);
		}
	}
}

void solve(){
	
	// int h,w;
	cin >> h >> w;
	int sx,sy,tx,ty;
	cin >> sx >> sy >> tx >> ty;
	sx--,sy--;tx--,ty--;
	for(int i =0;i<h;i++){
		for(int j =0;j<w;j++){
			cin >> maze[i][j];
			cache[i][j] = -1;
		}
	}	
	int comps = 0;
	for(int i =0;i<h;i++){
		for(int j =0;j<w;j++){
			if(pos(i,j) && cache[i][j]==-1){
				cout << i << " " << j << endl;
				dfs(i,j,comps);
				comps++;
			}
		}
	}

	// now I will add the edges

	for(int i =0;i<h;i++){
		for(int j=0;j<w;j++){
			for(int xx = -2;xx<3;xx++){
				for(int yy = -2;yy<3;yy++){
					int nx = i+xx,ny = j+yy;
					if(pos(nx,ny)&&pos(i,j)){
						if(cache[i][j]!=cache[nx][ny]){
							if(cache[i][j]>=0 && cache[nx][ny]>=0){
								// cout << i << " " << j<< " " << nx << " " << ny << endl;
								adj[cache[i][j]].insert(cache[nx][ny]);
								adj[cache[nx][ny]].insert(cache[i][j]);
							}
						}
					}
				}
			}
		}
	}

	// now just find the shortest path from x to y;
	// cout << "done" << endl;
	vis = vector<int> (comps,0);
	if(cache[tx][ty]==-1 || cache[sx][sy]==-1){
		cout << -1 << endl;return;
	}
	int s= cache[sx][sy];
	int t =cache[tx][ty];
	// dbg(adj); 
	if(s==t && s!=-1){

		cout << 0 << endl;
	}
	else{
		bfs(s,t,0);
		cout << (ans>=1e9?-1:ans) << endl;
		// cout << ans << endl;
	}

	// jst need to check dp[tx][ty]
}

int main(){
	fast;
	#ifndef ONLINE_JUDGE
		freopen("input.txt","r",stdin);
		freopen("output.txt","w",stdout);
	#endif

	// int t;cin >> t;while(t--)
		solve();
}