#include <bits/stdc++.h>

using namespace std;
ifstream f("car.in");
ofstream g("car.out");
const int oo = 1<<30;
int n,m,ip,jp,is,js,a[502][502][8];
int depI[8]={-1,-1,0,1,1, 1, 0,-1};
int depJ[8]={ 0, 1,1,1,0,-1,-1,-1};
int rotL[8]={7,0,1,2,3,4,5,6};
int rotR[8]={1,2,3,4,5,6,7,0};
queue<tuple<int,int,int,int>> q,qaux;
int main()
{
    f>>n>>m>>ip>>jp>>is>>js;

    for(int i=1; i<=n; i++)
        for(int j=1; j<=m; j++)
        {
            int x;
            f>>x;
            if(x==0)
                for(int k=0; k<8; k++)
                    a[i][j][k]=oo;
        }
    if(a[ip][jp][0]==0||a[is][js][0]==0)
    {
        g<<"-1\n";
        return 0;
    }
    if(is==ip&&js==jp)
    {
        g<<"0\n";
        return 0;
    }
    for(int k=0;k<8;k++)
    {
        a[ip][jp][k]=0;
        q.push(make_tuple(0,k,ip,jp));
    }
    while(q.size())
    {
        int cst,dir,i,j;
        tie(cst,dir,i,j)=q.front();
        q.pop();
        /// sunt intr-o anumita pozitie pozitionat pe o anumita directie si am ajuns acolo cu un anumit cost
        /// aseasta stare are cel mai mic cost disponibil
        /// din aceasta pozitie am 3 optiuni
            ///-> avansez fara sa fac curba => am acelasi cost
            ///-> fac 45 grade dreapta => cresc costul cu 1
            ///-> fac 45 grade stanga => cresc costul cu 1
        /// ajung intr-o noua stare pe care o iau in considerare doar daca imi imbunatateste costul
        /// VARIANTA 1 -> ma misc
        int I=i+depI[dir];
        int J=j+depJ[dir];
            if(cst<a[I][J][dir])
            {
                a[I][J][dir]=cst;
                q.push(make_tuple(cst,dir,I,J));
            }
        int DIR=rotL[dir];
            if(cst+1<a[i][j][DIR])
            {
                a[i][j][DIR]=cst+1;
                qaux.push(make_tuple(cst+1,DIR,i,j));
            }
        DIR=rotR[dir];
           if(cst+1<a[i][j][DIR])
            {
                a[i][j][DIR]=cst+1;
                qaux.push(make_tuple(cst+1,DIR,i,j));
            }
        if(q.size()==0)
        {
            while(qaux.size())
            {
                q.push(qaux.front());
                qaux.pop();
            }
        }

    }
    int sol=a[is][js][0];
    for(int k=1;k<8;k++)
        sol=min(sol,a[is][js][k]);
    if(sol==oo)
        sol=-1;
    g<<sol;
    return 0;
}