CodeForces1010A Fly

题目传送门:http://codeforces.com/problemset/problem/1010/A

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 1.51.5 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

Copy

2
12
11 8
7 5

output

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10.0000000000

input

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3
1
1 4 1
2 5 3

output

Copy

-1

input

Copy

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

output

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85.4800000000

中文题意：一共n个星球，主人公从地球出发，飞到最后一颗星球（第n颗），然后再从第n颗星球飞回地球（第一颗），在每个星球上降落和起飞所需要的燃料都不同，（假设星球间飞行不需要燃料）给出起飞或降落消耗燃料的计算公式，询问这趟旅程最少需要消耗多少燃料。

策略：显然飞船质量越小消耗的燃料越少，这样飞船开始的时候带的燃料就可以越少，这两种影响对答案的影响方向都是相同的，所以我们果断选择飞船质量小。最优策略是飞船最后在地球上降落时没有燃料剩余，用公式求出最后一次燃料质量，这样递归求解即可。

#include
using namespace std;
typedef double ld;
int n;
ld m;
const int N = 1005;
ld a[N], b[N];

ld digui(int pos, ld m)
{
    if(pos<=1)
        return m;
	if (a[pos] <= 1 || b[pos] <= 1)
	{
		puts("-1");
		return -1;
	}
	m += m / (a[pos] - 1);
	m += m / (b[pos] - 1);
	return digui(pos - 1, m);
}

int main()
{
	scanf("%d", &n);
	scanf("%lf", &m);
	for (int i = 1; i <= n; i++)
		scanf("%lf", &a[i]);
	for (int i = 1; i <= n; i++)
		scanf("%lf", &b[i]);
	if (b[1] <= 1)
	{
		puts("-1");
		return 0;
	}
	ld no = digui(n, m / (b[1] - 1) + m);
	if (no == -1)
		return 0;
	if (a[1] <= 1)
	{
		puts("-1");
		return 0;
	}
	no += no / (a[1] - 1);
	printf("%.8f\n", no - m);
	return 0;
}