Codeforces 701D. As Fast As Possible（二分）

D. As Fast As Possible

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the lengthl meters. Each of the pupils will go with the speed equal tov₁. To get to the excursion quickly, it was decided to rent a bus, which has seats fork people (it means that it can't fit more thank people at the same time) and the speed equal tov₂. In order to avoid seasick, each of the pupils want to get into the busno more than once.

Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.

Input

The first line of the input contains five positive integers n, l, v₁,v₂ andk (1 ≤ n ≤ 10 000,1 ≤ l ≤ 10⁹,1 ≤ v₁ < v₂ ≤ 10⁹,1 ≤ k ≤ n) — the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.

Output

Print the real number — the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed10^- 6.

Examples

Input

5 10 1 2 5

Output

5.0000000000

Input

3 6 1 2 1

Output

4.7142857143

Note

In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5.

题意：现在有n个人，路程长度为l，人行走速度为v1，有一辆车，每次能载k人，车行走速度为v2，每个人最多只能坐一次车，求n个人走完l的最少花费时间。

思路:每个人只能坐一次车，所以最后到达终点的人决定总时间，可以看出，用车每次载k个人，行走一定的距离，然后折返载下一趟人追上上一趟的人这种情

况能得出最优值。所以假设第一趟车载人时间为t1，所以两边的人相距v2*t1-v1*t1，假设第二趟车载人时间为t2，因为要追上上一趟，所以v2*t1-v1*t1+v1*t2=v2*t2，

所以t1=t2，也就是每人坐车的时间相等。

车载人的趟数：p=(n+k-1)/k。

设总时间为t，每人坐车的时间t1，则v1*(t-t1)+v2*t1=l，所以t1可以用t表示，设折返时间为t2，则(v2-v1)*t1=(v2+v1)*t2，而且t2*(p-1)+t1*p<=t。这样二分时间t就

可以求出结果。

#include <bits/stdc++.h>using namespace std;double v1, v2, l;int n, k;bool check(double t){    double t2 = (l-t*v1)/(v2-v1);    double t3 = (v2-v1)*t2/(v2+v1);    int p = (n+k-1)/k;    return t2*p+(p-1)*t3<=t;}int main(){    scanf("%d %lf %lf %lf %d", &n, &l, &v1, &v2, &k);    double ll = l/v2, r = l/v1, mid;    for(int i = 1;i <= 100000;i++){        mid = (ll+r)/2;        if(check(mid)) r = mid;        else ll = mid;    }    printf("%.7lf\n", r);}