Codeforces Round #499 (Div. 2) C. Fly(数学+思维模拟)

C. Fly

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

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.

input

2
12
11 8
7 5

output

10.0000000000

input

3
1
1 4 1
2 5 3

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 22 tons 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.

题意：

给你飞机初始重量，飞机要在所有机场起飞降落。一开始会携带一定重量的汽油，然后每次飞的过程会消耗掉相应汽油，而每个机场给的值是1吨汽油能让他飞多少。

求一开始最少携带多少汽油。

解题思路：

拿第一组样例来说

11 8
7 5

欲最小，回到起点时候肯定就耗光了汽油，而最后一个待过的机场是 7 机场，因为2*7=12+2,2吨汽油能让他飞14，恰好是飞机重量+耗油重量，依次回推2*8=14+2,16+4=5*4,20+2=11*2

得 x*(a[i]或者b[i])=s+x  ,s为下一站油和机身的和，x和具体某个机场的耗油量。

说是数学，其实就是证明下就能发现，无论a[i] b[i]用哪个顺序，最后耗油最小量都一样

再注意点细节，就ok了

#include <bits/stdc++.h>
using namespace std;
#define ll            long long
#define FastRead        ios_base::sync_with_stdio(0);cin.tie(0)

const int maxn=1e5+;
const int mod=1e9+;

int n,a[],b[];
double s;
int main()
{
    cin>>n>>s;
    int f=;
    for(int i=;i<=n;i++) {cin>>a[i];if(a[i]<=) f=;}
    for(int i=;i<=n;i++) {cin>>b[i];if(b[i]<=) f=;}
    if(f==) { puts("-1");return ; }
    double sum=,x;
    for(int i=;i<=n;i++)
    {
        x=s/(a[i]-),s+=x,sum+=x;
        x=s/(b[i]-),s+=x,sum+=x;
    }
    printf("%.10f\n",sum);
    return ;
}