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>
//#pragma GCC optimize(2)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<cstring>
#include<bitset>
#include<stack>
#include<time.h>



#define X first
#define Y second
#define PB push_back
#define MP make_pair
#define scd(a) scanf("%d",&a)
#define scdd(a,b) scanf("%d%d",&a,&b)
#define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define ALL(x) x.begin(),x.end()
#define sz(a) ((int)a.size())
#define getmid ((l+r)>>1)
#define mst(var,val) memset(var,val,sizeof(var))
#define IOS ios::sync_with_stdio(false);cin.tie(0)
#define lowbit(x) x&(-x)
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
#define ls rt<<1
#define rs rt<<1|1
using namespace std;
#ifdef local
#define dbg(args...) cout << #args << " -> ", err(args);
void err(){ cout << endl; }
template<typename T, typename... Args>
void err(T a, Args... args){ cout << a << ' '; err(args...); }
#else
#define dbg(args...)
#endif // local

typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <double, double> pdd;
const int inf=0x3f3f3f3f;
const long long INF=0x3f3f3f3f3f3f3f3fLL;
const double PI=acos(-1.0);
const long double eps=1e-8;
const int mod=1e9+7;
const int maxn=3e5+100;
const int N=2e5+1000;
const int M=500+10;
const ll mm=(1LL<<32);

template <class T>
inline void read(T &x)
{
    x = 0;
    char c = getchar();
    bool f = 0;
    for (; !isdigit(c); c = getchar())
        f ^= c == '-';
    for (; isdigit(c); c = getchar())
        x = x * 10 + (c ^ 48);
    x = f ? -x : x;
}

template <class T>
inline void write(T x)
{
    if (x < 0)
    {
        putchar('-');
        x = -x;
    }
    T y = 1;
    int len = 1;
    for (; y <= x / 10; y *= 10)
        ++len;
    for (; len; --len, x %= y, y /= 10)
        putchar(x / y + 48);
}
ll qpow(ll a,ll b,ll mod)
{
    ll ans=1;
    while(b)
    {
        if(b&1)
            ans=(ans*a)%mod;
        b>>=1;
        a=(a*a)%mod;
    }
    return ans;
}

struct node{
    int x,y,d;
    bool operator<(const node&b)const{
        return d>b.d;
    }
};

int dis[1010][1010];
char s[1010][1010];
int n,m,fs,ft,es,et;
bool is(int x,int y)
{
    return x>=1&&x<=n&&y>=1&&y<=m&&s[x][y]=='.';
}
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
int tx[]={-2,-2,-2,-2,-2,-1,-1,-1,-1,-1,0,0,0,0,1,1,1,1,1,2,2,2,2,2};
int ty[]={-2,-1,0,1,2,-2,-1,0,1,2,-2,-1,1,2,-2,-1,0,1,2,-2,-1,0,1,2};
void dij()
{
    priority_queue<node>q;
    dis[fs][ft]=0;
    q.push({fs,ft,0});
    while(!q.empty())
    {
        node tmp=q.top();q.pop();
        int x=tmp.x,y=tmp.y,d=tmp.d;
        if(d>dis[x][y]) continue;
        int nx,ny;
        for(int i=0;i<3;++i)
        {
            nx=x+dx[i];ny=y+dy[i];
            if(is(nx,ny)&&dis[nx][ny]>dis[x][y])
            {
                dis[nx][ny]=dis[x][y];
                q.push({nx,ny,dis[nx][ny]});
            }
        }
        for(int i=0;i<24;++i)
        {
            nx=x+tx[i];ny=y+ty[i];
            if(is(nx,ny)&&dis[nx][ny]>dis[x][y]+1)
            {
                dis[nx][ny]=dis[x][y]+1;
                q.push({nx,ny,dis[nx][ny]});
            }
        }
    }
}
int main()
{
#ifdef local
    freopen("in.txt","r",stdin);
#endif // local
    IOS;cout.tie(0);
    cin>>n>>m;
    cin>>fs>>ft;
    cin>>es>>et;
    rep1(i,n) cin>>(s[i]+1);
    mst(dis,inf);
    dij();
    if(dis[es][et]==inf) cout<<-1<<endl;
    else cout<<dis[es][et]<<endl;
    return 0;
}