With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked to carefully design the cheapest route to go.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 positive numbers: Cmax (<= 100), the maximum capacity of the tank; D (<=30000), the distance between Hangzhou and the destination city; Davg (<=20), the average distance per unit gas that the car can run; and N (<= 500), the total number of gas stations. Then N lines follow, each contains a pair of non-negative numbers: Pi, the unit gas price, and Di (<=D), the distance between this station and Hangzhou, for i=1,…N. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the cheapest price in a line, accurate up to 2 decimal places. It is assumed that the tank is empty at the beginning. If it is impossible to reach the destination, print “The maximum travel distance = X” where X is the maximum possible distance the car can run, accurate up to 2 decimal places.

Sample Input 1:

50 1300 12 8

6.00 1250

7.00 600

7.00 150

7.10 0

7.20 200

7.50 400

7.30 1000

6.85 300

Sample Output 1:

749.17

Sample Input 2:

50 1300 12 2

7.10 0

7.00 600

Sample Output 2:

The maximum travel distance = 1200.00

1 | #include "stdio.h" |

策略1：优先前往油价更低的加油站

策略2：在没有更低油价的加油站时，往油价尽可能低的加油站

合并在一起，就是在所有满油状态能到达的加油站中，选出油价最低的那个加油站，而一旦在枚举过程中找到了第一个油价低于当前油价的加油站，则退出循环，结束选择过程。