CF #499 div2 C. Fly(模拟)

output

standard output

Natasha is going to fly on a rocket to Mars and return to Earth. Also, on the way to Mars, she will land on n−2n−2 intermediate planets. Formally: we number all the planets from 11 to nn. 11 is Earth, nn is Mars. Natasha will make exactly nn flights: 1→2→…n→11→2→…n→1.

Flight from xx to yy consists of two phases: take-off from planet xx and landing to planet yy. This way, the overall itinerary of the trip will be: the 11-st planet →→ take-off from the 11-st planet →→ landing to the 22-nd planet →→ 22-nd planet →→ take-off from the 22-nd planet →→ …… →→ landing to the nn-th planet →→ the nn-th planet →→ take-off from the nn-th planet →→ landing to the 11-st planet →→ the 11-st planet.

The mass of the rocket together with all the useful cargo (but without fuel) is mm tons. However, Natasha does not know how much fuel to load into the rocket. Unfortunately, fuel can only be loaded on Earth, so if the rocket runs out of fuel on some other planet, Natasha will not be able to return home. Fuel is needed to take-off from each planet and to land to each planet. It is known that 11 ton of fuel can lift off aiai tons of rocket from the ii-th planet or to land bibi tons of rocket onto the ii-th planet.

For example, if the weight of rocket is 99 tons, weight of fuel is 33 tons and take-off coefficient is 88 (ai=8ai=8), then 1.51.5 tons of fuel will be burnt (since 1.5⋅8=9+31.5⋅8=9+3). The new weight of fuel after take-off will be 1.51.5tons.

Please note, that it is allowed to burn non-integral amount of fuel during take-off or landing, and the amount of initial fuel can be non-integral as well.

Help Natasha to calculate the minimum mass of fuel to load into the rocket. Note, that the rocket must spend fuel to carry both useful cargo and the fuel itself. However, it doesn't need to carry the fuel which has already been burnt. Assume, that the rocket takes off and lands instantly.

Input

The first line contains a single integer nn (2≤n≤10002≤n≤1000) — number of planets.

The second line contains the only integer mm (1≤m≤10001≤m≤1000) — weight of the payload.

The third line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤10001≤ai≤1000), where aiai is the number of tons, which can be lifted off by one ton of fuel.

The fourth line contains nn integers b1,b2,…,bnb1,b2,…,bn (1≤bi≤10001≤bi≤1000), where bibi is the number of tons, which can be landed by one ton of fuel.

It is guaranteed, that if Natasha can make a flight, then it takes no more than 109109 tons of fuel.

Output

If Natasha can fly to Mars through (n−2)(n−2) planets and return to Earth, print the minimum mass of fuel (in tons) that Natasha should take. Otherwise, print a single number −1−1.

It is guaranteed, that if Natasha can make a flight, then it takes no more than 109109 tons of fuel.

The answer will be considered correct if its absolute or relative error doesn't exceed 10−610−6. Formally, let your answer be pp, and the jury's answer be qq. Your answer is considered correct if |p−q|max(1,|q|)≤10−6|p−q|max(1,|q|)≤10−6.

Examples

input

output

10.0000000000

input

output

-1

input

6
2
4 6 3 3 5 6
2 6 3 6 5 3

output

85.4800000000

Note

Let's consider the first example.

Initially, the mass of a rocket with fuel is 2222 tons.

At take-off from Earth one ton of fuel can lift off 1111 tons of cargo, so to lift off 2222 tons you need to burn 22tons of fuel. Remaining weight of the rocket with fuel is 2020 tons.
During landing on Mars, one ton of fuel can land 55 tons of cargo, so for landing 2020 tons you will need to burn 44 tons of fuel. There will be 1616 tons of the rocket with fuel remaining.
While taking off from Mars, one ton of fuel can raise 88 tons of cargo, so to lift off 1616 tons you will need to burn 22 tons of fuel. There will be 1414 tons of rocket with fuel after that.
During landing on Earth, one ton of fuel can land 77 tons of cargo, so for landing 1414 tons you will need to burn 22 tons of fuel. Remaining weight is 1212 tons, that is, a rocket without any fuel.

In the second case, the rocket will not be able even to take off from Earth.

题目大意：第一行给出一个数n，代表有n个点，第二行给出火箭的自重，第三行给出每个点去时每吨需要的燃料，第四行给出每个点返程时每吨需要的燃料。求出发时携带的最小燃料数。

思路：读不懂题真的是一个大问题，特别特别难受。该题是一个模拟题，要求最小携带燃料数量，那么最后一站到达时燃料数量就一定为0；那么上一站的燃料数就一定是(当前重量+燃料数)/每吨需要的燃料，所以x=(x+s)/a[i],所以s=x*a[i]-x，同理，b[i]做相同处理。至于为什么不考虑到达的顺序问题，因为无论顺序如何，所需要的最小燃料数都是一样的。(应该是移位后虽然一边变大了，同时另一边就变小了)

代码如下:

#include<bits/stdc++.h>
using namespace std;
int n,a[1005],b[1005];
double s;
int main()
{
    cin>>n>>s;
    int fag=1;
    for(int i=1;i<=n;i++) 
	{
	    cin>>a[i];
		if(a[i]<=1)//若等于1，消耗与载重1:1:(飞到了就耗完燃料) 
		fag=0;
	}
    for(int i=1;i<=n;i++) 
	{
	    cin>>b[i];
		if(b[i]<=1) 
		fag=0;
	}
    if(fag==0) 
	{ 
	    puts("-1");
		return 0;
    }
    double sum=0,x;
    sort(a,a+n);
    sort(b,b+n);
    for(int i=1;i<=n;i++)
    {
        x=s/(a[i]-1),s+=x,sum+=x;//cout<<s<<"*"<<sum<<endl; 
        x=s/(b[i]-1),s+=x,sum+=x;//cout<<s<<"*"<<sum<<endl;
    }
    printf("%.10f\n",sum);
    return 0;
}

CF #499 div2 C. Fly(模拟)

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