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>
using namespace std;

using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;
using LL = long long;
using VL = vector<LL>;
using VVL = vector<VL>;
using PLL = pair<LL, LL>;
using VS = vector<string>;

#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define FF first
#define SS second
template<class S, class T>
istream& operator>>(istream& is, pair<S,T>& p){
  return is >> p.FF >> p.SS;
}
template<class S, class T>
ostream& operator<<(ostream& os, const pair<S,T>& p){
  return os << p.FF << " " << p.SS;
}
template<class T>
void maxi(T& x, T y){
  if(x < y) x = y;
}
template<class T>
void mini(T& x, T y){
  if(x > y) x = y;
}


const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD = 1e9+7;

const int MAX = 1<<20;
LL N, M;
long double num[MAX], prob[MAX];

using LLL = __int128;

LLL gcd(LLL x, LLL y){
  if(y == 0) return x;
  return gcd(y, x%y);
}
LLL lcm(LLL x, LLL y){
  return x / gcd(x, y) * y;
}

int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  cin >> N >> M;
  VL xs(N);
  vector<long double> ps(N);
  REP(i,N)
	cin >> xs[i];
  REP(i,N){
	cin  >> ps[i];
	ps[i] /= 100.;
  }

  FOR(b,1,1<<N){
	LLL l = 1;
	int bc = 0;
	REP(i,N){
	  if(b>>i&1){
		if(l <= M)
		  l = lcm(l, xs[i]);
		++bc;
	  }
	  num[b] = (bc%2*2-1) * M / l;
	}
  }

  for(int i=0;i<N;++i){
	for(int b=0;b<(1<<N);++b){
	  if((b>>i&1) == 1)
		num[b] += num[b^(1<<i)];
	}
  }

  long double ans = 0.;
  REP(b,1<<N){
	prob[b] = 1;
	REP(i,N){
	  prob[b] *= (b>>i&1? ps[i]: 1.-ps[i]);
	}
	ans += prob[b] * num[b];
  }

  cout << fixed << setprecision(9) << ans << endl;
  
  return 0;
}