Problem: Given n integers a1, a2,…, an and n integers p1, p2,…, pn, integer m. The operation of selecting the kth integer ak with a probability of pk [%] is performed for each k (1 ≤ k ≤ n), and 0 or more and n or less integers are selected. Find the expected number of integers between 1 and m that are divisible by at least one of the selected integers.

Constraints

* 1 ≤ n ≤ 20

* 1 ≤ m ≤ 1018

* 1 ≤ ak ≤ 1018 (1 ≤ k ≤ n)

* 1 ≤ pk ≤ 99 (1 ≤ k ≤ n)

Input

The input is given in the following format.

> n m
> a1 a2… an
> p1 p2… pn
>

Output

Print the solution to the problem on one line. The output must not have an absolute or relative error of more than 0.0000001 (= 10-7).

Examples

Input

2 15
3 5
50 50


Output

3.75


Input

4 100
2 3 5 7
80 60 40 20


Output

57.352


Input

4 210
2 3 5 7
80 60 40 20


Output

119.9136

Solution (in CPP):

Solution:
#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
//#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<bitset>
#include<stack>
#include<memory>
using namespace std;
//using namespace std::chrono;
typedef long long int llint;
typedef double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase

/*
cout<<fixed<<setprecision(20);
cin.tie(0);
ios::sync_with_stdio(false);
*/
const llint mod=1000000007;
const llint big=2.19e15+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-7;
template <class T,class U>bool mineq(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool maxeq(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}

int main(void){
	cout<<fixed<<setprecision(20);
	cin.tie(0);
	ios::sync_with_stdio(false);
	llint n,m,i,bi,j;cin>>n>>m;
	static llint a[20]={0};
	static double p[20]={0};
	for(i=0;i<n;i++){cin>>a[i];}
	for(i=0;i<n;i++){cin>>p[i];p[i]/=100;}
	static double dp[1<<20];
	dp[0]=0;
	for(bi=1;bi<(1<<n);bi++){
		llint alc=1;
		bool flow=false;
		int ki=0;
		for(i=0;i<n;i++){
			if(!(bi&(1<<i))){continue;}
			ki++;
			llint gc=gcd(alc,a[i]);
			double miru=((double)(alc))*a[i]/gc;
			if(miru>m*1.5){flow=true;break;}
			alc/=gc;alc*=a[i];
		}
		if(flow){continue;}
		dp[bi]=m/alc;
		if(ki%2==0){dp[bi]*=-1;}
	}
	for(i=0;i<n;i++){
		for(bi=0;bi<(1<<n);bi++){
			if(bi&(1<<i)){continue;}
			dp[bi+(1<<i)]+=dp[bi];
		}
	}
	double ans=0;
	for(bi=0;bi<(1<<n);bi++){
		double gen=dp[bi];
		for(i=0;i<n;i++){
			if(bi&(1<<i)){gen*=p[i];}
			else{gen*=1-p[i];}
		}
		ans+=gen;
	}
	cout<<ans<<endl;
	return 0;
}