B. Number Busters

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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).

Output

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.

Examples

input

4 2 3 1 6

output

input

4 2 3 1 7

output

input

1 2 3 2 6

output

input

1 1 2 1 1

output

当时应该仔细分析一下这个题目的。

只有b>=x的时候这个时候c才开始追a。

而当b<x的时候c与a的差并不会发生变化。

所以当c与a追平的时候一定是在上面的轮回。

于是我们可以有式子：

b - x*(c-a) + (w-x)*k >= x - x;

也就是最极端的情况就是b刚好变成0的时候c追上a。

然后可以解k的不等式，这时候k要取整数，所以应该是向上取整。

最后加上一个c-a就ok了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include <map>
using namespace std;
const int MAXN = 2333;
const int inf = 1e9;
long long a,b,w,x,c;
int main()
{scanf("%I64d%I64d%I64d%I64d%I64d",&a,&b,&w,&x,&c);double k = (x*(c-a) - b)*1.0/(w-x);long long x = ceil(k);long long ans = c - a + (x>0?x:0);printf("%I64d\n",ans);return 0;
}

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