#include <iostream>

using namespace std;

int x=0,y=0,a,b,c,j=0,i=0,k=0,l=0,counter=0,maks=0,z=0,g=0,zet=0,countertwo=x,maksz=0;

int main()
{
    cin >> y >> x >> a;
    int ruchpotablicy[x*y];
    for(i=0; i<=ruchpotablicy[x*y]; i++)
    {
        ruchpotablicy[i]=0;
    }
    for(k=0; k<a; k++)
    {
        cin >> b >> c;
        b--;
        c--;
        ruchpotablicy[b*x+c]=1;
    }
    for(j=0; j<x*y; j++)
    {
        if((j%x==0)&&(j!=0))
        {
            cout << "\n";
        }
        cout << ruchpotablicy[j];
    }
    for(j=0; j<x*y; j++)
    {
        if(ruchpotablicy[counter+j]==1)
        {
            z++;
        }
        if(z>maks)
        {
            maks=z;
        }
        if((j%x==0)&&(j!=0))
        {
            counter++;
            z=0;
        }
    }
    z=0;
    counter=0;
    for(zet=0; zet<x*y+x; zet++)
    {
        if(ruchpotablicy[counter+g]==1)
        {
            z++;
        }
        counter=counter+x;
        if(z>maks)
        {
            maks=z;
        }
        if(counter>x*y)
        {
            counter=0;
            g++;
            z=0;
        }
    }
    z=0;
    for(zet=0; zet<x*y+x; zet++)
    {
        if((ruchpotablicy[countertwo]==1)&&(ruchpotablicy[countertwo-(x-1)]))
        {
            z++;
        } else if((ruchpotablicy[countertwo]==1)&&(ruchpotablicy[countertwo-(x-1)])) maksz++;
        else
        {
            maksz=z;
            z=0;
        }
        countertwo++;
    }
    if(z>maksz)
    {
        maks=maksz;
    }
    countertwo=0;
    z=0;
    for(zet=0; zet<x*y+x; zet++)
    {
        if((ruchpotablicy[countertwo]==1)&&(ruchpotablicy[countertwo-(x+1)]))
        {
            z++;
        } else if((ruchpotablicy[countertwo]==1)&&(ruchpotablicy[countertwo-(x+1)])) maksz++;
        else
        {
            maksz=z;
            z=0;
        }
        countertwo++;
    }
    if(z>maksz)
    {
        maks=maksz;
    }
    cout << "\n" << maks;
    /*for(l=0; l<a; l++)
    {

    }*/
    // x to lewo i prawo
    // y to gora i dol
}