CF382B.Number Busters

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 < w, 0 < x < 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 < 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.

亚瑟和亚历山大是数字破坏者。今天他们进行了一场竞赛。

亚瑟选取了四个整数 $a$、$b$、$w$、$x$（满足 $0 \leq b < w$，$0 < x < w$），而亚历山大选取了整数 $c$。亚瑟和亚历山大采用不同的数字破坏方法。亚历山大只是一个普通人：每秒他将自己的数减去 1，即执行赋值操作：$c = c - 1$。亚瑟则是一位复杂操作者：每秒他执行如下复杂操作：若 $b \geq x$，则执行赋值 $b = b - x$；若 $b < x$，则连续执行两个赋值操作：$a = a - 1$；$b = w - (x - b)$。

已知初始数值 $a$、$b$、$w$、$x$、$c$。请确定在两人同时开始执行操作的情况下，亚历山大何时超越亚瑟。我们定义“亚历山大超越亚瑟”为满足 $c \leq a$ 的时刻。

The first line contains integers a, b, w, x, c (1 ≤ a ≤ 2·109, 1 ≤ w ≤ 1000, 0 ≤ b < w, 0 < x < w, 1 ≤ c ≤ 2·109).

第一行包含整数 $a$、$b$、$w$、$x$、$c$（$1 ≤ a ≤ 2·10^9$，$1 ≤ w ≤ 1000$，$0 ≤ b < w$，$0 < x < w$，$1 ≤ c ≤ 2·10^9$）。

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.

输出一个整数——亚历山大超越亚瑟所需的最短时间（单位：秒）。可以证明，在本题的约束条件下，所述情况总是会发生。