Description

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases.

The input is terminated by a line containing pair of zeros

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Analysis

1. 给出 n 个岛和半径 d，岛的 y 坐标不小于 d，则必然失败。
2. 求出每个岛和 x 的两(一)个交点，则有 n 个区间，转化为区间贪心选择问题。
3. 将每个区间按照左端点的顺序增序排列。若下一个区间的右端点 < 此时的右端点，更新。否则若下一个区间的左端点大于此时的右端点，更新，答案+1。
4. 几何问题：每一个岛为圆心，d 为半径画的圆与 x 轴两个交点的区间，其中任意一点为圆心，d 为半径都可以到岛。因为任意一点，都在岛为圆心的园内，所以岛圆心的距离必然小于 d。
# POJ