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 <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<29;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

ll gcd(ll a,ll b){
	if(!b) return a;
	return gcd(b,a%b);
}

ll n,m;
vl a,b,dp[2];

int main(){
	cin>>n>>m;
	a=b=vl(n);
	ll N=1<<n;
	dp[0]=dp[1]=vl(N);
	for(auto &i:a) cin>>i;
	for(auto &i:b) cin>>i;
	for(int i=1;i<N;i++){
		ll t=1;
		for(int j=0;j<n;j++) if(i&1<<j){
			ll g=gcd(t,a[j]);
			if(a[j]/g<=(m+t-1)/t) t*=a[j]/g;
			else t=m+1;
			if(t>m) break;
		}
		int id=0;
		for(int j=0;j<n;j++) if(i&1<<j) id^=1;
		dp[id][i]=m/t;
	}
	for(int i=0;i<n;i++){
		vl DP[2];
		DP[0]=DP[1]=vl(N);
		for(int j=0;j<N;j++){
			DP[0][j]=dp[0][j]+(j&1<<i?dp[0][j-(1<<i)]:0);
			DP[1][j]=dp[1][j]+(j&1<<i?dp[1][j-(1<<i)]:0);
		}
		swap(dp,DP);
	}
	double res=0;
	for(int i=1;i<N;i++){
		double p=1;
		for(int j=0;j<n;j++){
			if(i&1<<j) p*=1.0/100*b[j];
			else p*=1.0/100*(100-b[j]);
		}
		res+=p*(dp[1][i]-dp[0][i]);
	}
	cout<<syosu(9)<<res<<endl;
}