// 可以选择二分查找

 #include <iostream>

 #include <algorithm>

 #include <cstdio>

 #include <cstring>

 using namespace std;

 typedef long long LL;

 const int Max =  + ;

 int num[Max];

 //int ux[Max], dx[Max];

 struct Point

 {

     LL ux, dx;

 };

 LL Cross(LL x1, LL y1, LL x2, LL y2)

 {

     return x1 * y2 - y1 * x2;

 }

 int main()

 {

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

     while (scanf("%d", &n) != EOF && n)

     {

         scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);

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

             scanf("%d%d", &ux[i], &dx[i]);

         ux[n + ] = x2;  // 把最后边界也算进去

         dx[n + ] = x2;

         memset(num, , sizeof(num));

         int x, y, l, r, mid;

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

         {

             scanf("%d%d", &x, &y);

             if (!(x >= x1 && x <= x2 && y >= y2 && y <= y1)) // 不在矩形内

                 continue;

             l = , r = n + ;

             while (l < r)

             {

                 mid = (l + r) / ;

                 if (Cross(x - dx[mid], y - y2, ux[mid] - dx[mid], y1 - y2) > )  // 如果 大于 0 说明 点在直线 右侧，所以改变左区间

                     l = mid + ;

                 else

                     r = mid; // r始终是满足条件的

             }

             num[r - ]++;

             /*

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

             {

                 if ( Cross(x - dx[j], y - y2, ux[j] - dx[j], y1 - y2) <= 0)

                 {

                     num[j - 1]++;

                     break;

                 }

             }

             */

         }

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

             printf("%d: %d\n", i, num[i]);

         printf("\n");

     }

     return ;

 }