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>
using namespace std;
typedef long long int llint;
typedef long 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<<setprecision(20)
const llint mod=1000000007;
const llint big=2.19e15+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-15;
template <class T,class U>void mineq(T& a,U b){if(a>b){a=b;}}
template <class T,class U>void maxeq(T& a,U b){if(a<b){a=b;}}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint 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());}
int LBI(vector<llint>&ar,llint in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
int UBI(vector<llint>&ar,llint in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}
int main(void){
	llint n,m,i;cin>>n>>m;
	//包除
	lldo ans=0;
	llint aa[20]={0};
	lldo pp[20]={0};
	for(i=0;i<n;i++){cin>>aa[i];}
	for(i=0;i<n;i++){cin>>pp[i];pp[i]/=100;}
	for(int bi=1;bi<(1<<n);bi++){
		llint gen=1;
		lldo per=1;
		bool ok=true;
		int guki=0;
		for(i=0;i<n;i++){
			if(!(bi&(1<<i))){continue;}
			guki++;
			per*=pp[i];
			llint gc=gcd(gen,aa[i]);
			if(((lldo)gen/gc *aa[i])>2e18){ok=false;break;}
			gen=gen/gc *aa[i];
		}
		if(!ok){continue;}
		llint num=m/gen;
		if(guki%2==1){ans+=per*num;}
		else{ans-=per*num;}
	}
	cout<<setprecision(20)<<ans<<endl;
	return 0;
}