//第一发：莽写（利用ToLefttest）

#include<iostream>

#include<algorithm>

#include<cstring>

using namespace std;

struct Point{

int x,y;

}edge[5005],point[5005];

int a[5005];

bool ToLeft(int a,int b,int c,int d,int e,int f)

{

    if((a*d+b*e+c*f-d*e-b*c-a*f)<0)return true;

    else return false;

}

int main ()

{

    int n,m;

    while(cin>>n,n)

    {

        cin>>m;

        memset(a,0,sizeof(a));

        cin>>edge[0].x>>edge[0].y>>point[0].x>>point[0].y;

        for(int i=1;i<=n;i++)

            cin>>edge[i].x>>edge[i].y;

        for(int i=1;i<=m;i++)

            cin>>point[i].x>>point[i].y;

        for(int i=1;i<=n;i++)

            for(int j=1;j<=m;j++)

            if(ToLeft(edge[i].x,edge[0].y,edge[i].y,point[0].y,point[j].x,point[j].y))

                a[i]++;

        for(int i=0;i<n;i++)

            cout<<i<<':'<<' '<<a[i+1]-a[i]<<endl;

        cout<<n<<':'<<' '<<m-a[n]<<endl<<endl;

    }

    return 0;

}

#include<cstring >

#include<iostream>

using namespace std;

//第二发：参考kuangbin（利用二分+叉积）

struct Point {

    int x,y;

    Point (){};

    Point(int _x,int _y)

    {

        x=_x,y=_y;

    }

    Point operator -(const Point &b)const

    {

        return Point(x-b.x,y-b.y);

    }

    int operator *(const Point &b)const

    {

        return x*b.x+y*b.y;

    }

    int operator ^(const Point &b)const

    {

        return x*b.y-y*b.x;

    }

};

struct Line{

    Point s,e;

    Line(){};

    Line(Point _s,Point _e)

    {

        s=_s,e=_e;

    }

};

int xmult(Point p0,Point p1,Point p2)//求P0p1和 p0p2 的叉积

{

    return (p1-p0)^(p2-p0);

}

const int  MAXN=5050;

Line line[MAXN];

int ans[MAXN];

int main ()

{

    int n,m,x1,y1,x2,y2;

    while(cin>>n,n)

    {

        cin>>m>>x1>>y1>>x2>>y2;

        int U,L;

        for(int i=0;i<n;i++)

        {

            cin>>U>>L;

            line[i]=Line(Point(U,y1),Point(L,y2));

        }

        line[n]=Line(Point(x2,y1),Point(x2,y2));

        int x,y;

        Point p;

        memset(ans,0,sizeof(ans));

        while(m--)

        {

            cin>>x>>y;

            p=Point(x,y);

            int l=0,r=n;

            while(l<r)

            {

                int mid=(l+r)/2;

                if(xmult(p,line[mid].e,line[mid].s)>0)

                    r=mid;

                else

                    l=mid+1;

            }

            ans[l]++;

        }

        for(int i=0;i<=n;i++)

            cout<<i<<": "<<ans[i]<<endl;

            cout<<endl;

    }

    return 0;

}