Codeforces Round #499 (Div. 2) C Fly题解

题目

http://codeforces.com/contest/1011/problem/C

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 aiaitons 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 1.51.5 tons.

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

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
题意：
有n个星球，从1飞到n，路径为1->2->3->……->n->1,数组a中a[i]表示从第i个星球起飞一吨燃料可以支持多少重量起飞(飞船重量m和燃料重量总和为总重量)，

数组b中b[i]表示降落到第i个星球一吨燃料可以支持多少重量降落(飞船重量m和燃料重量总和为总重量)。
思路：
反向思考，设当前操作需要燃料x吨，除当前燃料所需外飞船总重量为y,当前1吨燃料可以支持的重量为t，则（x+y）/t=x;
所以x=y/(t-1);再进行前一步操作时将y加上x，由此式子也可看出t必须大于1，否则无法完成。
代码：

#include<bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pqueue priority_queue
#define NEW(a,b) memset(a,b,sizeof(a))
const double pi=4.0*atan(1.0);
const double e=exp(1.0);
const int maxn=3e6+;
typedef long long LL;
typedef unsigned long long ULL;
//typedef pair<LL,LL> P;
const LL mod=1e9+;
using namespace std;
int a[];
int b[];
int c[];
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(int i=;i<n;i++){
        scanf("%d",&b[i]);
    }
    int cnt=;
    for(int i=;i<n;i++){
        c[cnt++]=a[i];
        if(i!=n-){
            c[cnt++]=b[i+];
        }
        else{
            c[cnt++]=b[];
        }
    }//记录操作顺序.
    double g=double(m);
    double k;
　　 //反向操作.
    for(int i=cnt-;i>=;i--){
        if(c[i]<=){
            printf("-1\n");
            return ;
        }
        k=g/(double(c[i])-1.0);
        g+=k;
    }
    double ans=g-(double)m;
    printf("%.10lf\n",ans);
}