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: Time Limit Exceeded)


/*
LANG:C++
PROB:

*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
#include <iostream>
// #define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)


using namespace __gnu_pbds;
using namespace std;


typedef tree<int, null_type, less<int>, rb_tree_tag,
            tree_order_statistics_node_update>
    ordered_set;

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;
typedef unsigned long long ui64;

template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }
const int mod=1000000007;
  i64 poww(i64 a, i64 b)
  {
    a%=mod;
    i64 res=1;
    while(b)
      {
if(b&1)res=res*a%mod;

      a=a*a%mod;
      b/=2;

      }
      return res;
  }
  i64 inv(i64 a)
  {
    return poww(a,mod-2);
  }
  void solve()
  {

int h,w,sx,sy,ex,ey;
cin>>h>>w>>sx>>sy>>ex>>ey;
vector<vector<char>>grid(h+1,vector<char>(w+1));
for1(i,h)for1(j,w)cin>>grid[i][j];
queue<pair<pii,int>>q;	
q.push({{sx,sy},0});

int ans=INT_MAX;
vector<vector<bool>>vis(h+1,vector<bool>(w+1,false));
vector<vector<int>>dis(h+1,vector<int>(w+1,1e6));
dis[sx][sy]=0;
while(q.size())
{

	pair<pii,int>pp=q.front();
	q.pop();

	
	int hc=pp.fi.fi,wc=pp.fi.se;
	for(int i=-2;i<=2;i++)
	{
		for(int j=-2;j<=2;j++)
		{
int hn=hc+i,wn=wc+j;
if(i==0&&j==0)continue;

if(hn>=1&&hn<=h&&wn>=1&&wn<=w&&grid[hn][wn]!='#')
{
	int ex=1;
	if((i==0&&j==1)||(i==1&&j==0)||(i==-1&&j==0)||(i==0&&j==-1))ex=0;
     if(dis[hn][wn]>pp.se+ex)
     {
	q.push({{hn,wn},pp.se+ex});


dis[hn][wn]=pp.se+ex;
}
}
		}
	}

}



ans=dis[ex][ey];

if(ans==1e6)ans=-1;

cout<<ans;
  }
  
  int main()
  {
ios::sync_with_stdio(false);
    cin.tie(nullptr);
    #ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
    #endif
int t=1;
while(t--)
{
  solve();
}

    return 0;
  }