Number Busters

两个人玩游戏，他们有 $a$，$b$，$w$，$x$，$c$ 五个数，每秒可执行一次操作，要使最终结果为 $c\le a$。操作如下：如果 $b\ge x$，则 $b=b-x$，同时 $c=c-1$；如果 $b<x$，则 $a=a-1$，$c=c-1$，$b=w-(x-b)$。求 $c\le a$ 时已经走过的秒数。

Arthur and Alexander are number busters. Today they've got a competition. Arthur took a group of four integers $ a,b,w,x $ $ (0<=b&lt;w,0&lt;x&lt;w) $ and Alexander took integer $ с $ . Arthur and Alexander use distinct approaches to number bustings. Alexander is just a regular guy. Each second, he subtracts one from his number. In other words, he performs the assignment: $ c=c-1 $ . Arthur is a sophisticated guy. Each second Arthur performs a complex operation, described as follows: if $ b>=x $ , perform the assignment $ b=b-x $ , if $ b&lt;x $ , then perform two consecutive assignments $ a=a-1; b=w-(x-b) $ . You've got numbers $ a,b,w,x,c $ . Determine when Alexander gets ahead of Arthur if both guys start performing the operations at the same time. Assume that Alexander got ahead of Arthur if $ c<=a $ .

The first line contains integers $ a,b,w,x,c $ $ (1<=a<=2·10^{9},1<=w<=1000,0<=b&lt;w,0&lt;x&lt;w,1<=c<=2·10^{9}) $ .

Print a single integer — the minimum time in seconds Alexander needs to get ahead of Arthur. You can prove that the described situation always occurs within the problem's limits.

4 2 3 1 6

2

4 2 3 1 7

4

1 2 3 2 6

13

1 1 2 1 1

0