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
−
2
intermediate planets. Formally: we number all the planets from
1
to
n
.
1
is Earth,
n
is Mars. Natasha will make exactly
n
flights:
1
→
2
→
…
n
→
1
.

Flight from
x
to
y
consists of two phases: take-off from planet
x
and landing to planet
y
. This way, the overall itinerary of the trip will be: the
1
-st planet
→
take-off from the
1
-st planet
→
landing to the
2
-nd planet
→

2
-nd planet
→
take-off from the
2
-nd planet
→

…

→
landing to the
n
-th planet
→
the
n
-th planet
→
take-off from the
n
-th planet
→
landing to the
1
-st planet
→
the
1
-st planet.

The mass of the rocket together with all the useful cargo (but without fuel) is
m
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
1
ton of fuel can lift off
a
i
tons of rocket from the
i
-th planet or to land
b
i
tons of rocket onto the
i
-th planet.

For example, if the weight of rocket is
9
tons, weight of fuel is
3
tons and take-off coefficient is
8
(
a

i

8
), then
1.5
tons of fuel will be burnt (since
1.5
⋅

8

9
+
3
). The new weight of fuel after take-off will be
1.5
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
n
(
2
≤
n
≤
1000
) — number of planets.

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

The third line contains
n
integers
a
1
,
a
2
,
…
,
a
n
(
1
≤
a
i
≤
1000
), where
a
i
is the number of tons, which can be lifted off by one ton of fuel.

The fourth line contains
n
integers
b
1
,
b
2
,
…
,
b
n
(
1
≤
b
i
≤
1000
), where
b
i
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
10
9
tons of fuel.

Output
If Natasha can fly to Mars through
(
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
.

It is guaranteed, that if Natasha can make a flight, then it takes no more than
10
9
tons of fuel.

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

Examples
inputCopy
2
12
11 8
7 5
outputCopy
10.0000000000
inputCopy
3
1
1 4 1
2 5 3
outputCopy
-1
inputCopy
6
2
4 6 3 3 5 6
2 6 3 6 5 3
outputCopy
85.4800000000
Note
Let’s consider the first example.

Initially, the mass of a rocket with fuel is
22
tons.

At take-off from Earth one ton of fuel can lift off
11
tons of cargo, so to lift off
22
tons you need to burn
2
tons of fuel. Remaining weight of the rocket with fuel is
20
tons.
During landing on Mars, one ton of fuel can land
5
tons of cargo, so for landing
20
tons you will need to burn
4
tons of fuel. There will be
16
tons of the rocket with fuel remaining.
While taking off from Mars, one ton of fuel can raise
8
tons of cargo, so to lift off
16
tons you will need to burn
2
tons of fuel. There will be
14
tons of rocket with fuel after that.
During landing on Earth, one ton of fuel can land
7
tons of cargo, so for landing
14
tons you will need to burn
2
tons of fuel. Remaining weight is
12
tons, that is, a rocket without any fuel.
In the second case, the rocket will not be able even to take off from Earth.

自然是最后燃料全部用完是最优解，那么推导（乱搞）就好了；
( m+x )* ( 1- 1/a1 )* ( 1- 1/a2 )…( 1-1/an )( 1-1/b1 )( 1-1/b2 )…( 1-1/bn )== m；
当然 其中的 ai,bi 有一个<=1，自然就是无解了；
将上述方程解出 x 即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>

typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 200005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-5
#define pll pair<ll,ll>



ll quickpow(ll a, ll b) {
    ll ans = 1;
    while (b > 0) {
        if (b % 2)ans = ans * a;
        b = b / 2;
        a = a * a;
    }
    return ans;
}

int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a%b);
}



double a[1005], b[1004];

int main()
{
    ios::sync_with_stdio(false);
    int n, m;
    cin >> n >> m;
    int i, j;
    int flag = 0;
    for (i = 0; i < n; i++) {
        cin >> a[i];
        if (a[i] <= 1.0)flag = 1;
    }
    for (i = 0; i < n; i++) {
        cin >> b[i];
        if (b[i] <= 1.0)flag = 1;
    }
    if (flag)cout << -1 << endl;
    else {
        double ans = 1.0;
        for (i = 0; i < n; i++) {
            ans = 1.0*ans*(a[i] / (a[i] - 1.0))*(b[i] / (b[i] - 1.0));
        }
        ans = (ans - 1.0)*m;
        printf("%.10f\n", 1.000000*ans);

    }

}