1 //2017-08-30
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 const int N = 5010;
10 
11 struct Point{
12     int x, y;
13     Point(){}
14     Point(int _x, int _y):x(_x), y(_y){}
15     //a-b 表示向量 ba
16     Point operator- (const Point &b) const {
17         return Point(x-b.x, y-b.y);
18     }
19     //向量叉积
20     int operator* (const Point &b) const {
21         return x*b.y - y*b.x;
22     }
23 }A, B;
24 
25 int ans[N], U[N], L[N];
26 int n, m;
27 
28 bool check(int id, int x, int y){
29     Point a(L[id], B.y);
30     Point b(U[id], A.y);
31     Point c(x, y);
32     //令I = 向量ab 叉乘 向量 bc，若I为正，点c在向量ab的左侧（沿向量方向看）；为负则在右侧
33     return ((c-a)*(b-a)) > 0;
34 }
35 
36 int get_position(int x, int y){
37     int l = 0, r = n+1, mid, ans;
38     while(l <= r){
39         mid = (l+r)>>1;
40         if(check(mid, x, y)){
41             ans = mid;
42             l = mid+1;
43         }else r = mid-1;
44     }
45     return ans;
46 }
47 
48 int main()
49 {
50     std::ios::sync_with_stdio(false);
51     //freopen("inputA.txt", "r", stdin);
52     while(cin>>n && n){
53         cin>>m>>A.x>>A.y>>B.x>>B.y;
54         U[0] = L[0] = A.x;
55         U[n+1] = L[n+1] = B.x;
56         for(int i = 1; i <= n; i++)
57           cin>>U[i]>>L[i];
58         memset(ans, 0, sizeof(ans));
59         int x, y;
60         for(int i = 0; i < m; i++){
61             cin>>x>>y;
62             ans[get_position(x, y)]++;
63         }
64         for(int i = 0; i <= n; i++)
65           cout<<i<<": "<<ans[i]<<endl;
66         cout<<endl;
67     }
68 
69     return 0;
70 }