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 rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
//#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//---------------------------------------------------------------------------------------------------
template<int MOD> struct ModInt {
    static const int Mod = MOD; unsigned x; ModInt() : x(0) { }
    ModInt(signed sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    ModInt(signed long long sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    int get() const { return (int)x; }
    ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
    ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    ModInt inverse() const { long long a = x, b = MOD, u = 1, v = 0;
        while (b) { long long t = a / b; a -= t * b; std::swap(a, b); u -= t * v; std::swap(u, v); }
        return ModInt(u); }
    bool operator==(ModInt that) const { return x == that.x; }
    bool operator!=(ModInt that) const { return x != that.x; }
    ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
template<int MOD> ostream& operator<<(ostream& st, const ModInt<MOD> a) { st << a.get(); return st; };
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
    ModInt<MOD> r = 1; while (k) { if (k & 1) r *= a; a *= a; k >>= 1; } return r; }
template<typename T, int FAC_MAX> struct Comb { vector<T> fac, ifac;
    Comb(){fac.resize(FAC_MAX,1);ifac.resize(FAC_MAX,1);rep(i,1,FAC_MAX)fac[i]=fac[i-1]*i;
        ifac[FAC_MAX-1]=T(1)/fac[FAC_MAX-1];rrep(i,FAC_MAX-2,1)ifac[i]=ifac[i+1]*T(i+1);}
    T aPb(int a, int b) { if (b < 0 || a < b) return T(0); return fac[a] * ifac[a - b]; }
    T aCb(int a, int b) { if (b < 0 || a < b) return T(0); return fac[a] * ifac[a - b] * ifac[b]; }
    T nHk(int n, int k) { if (n == 0 && k == 0) return T(1); if (n <= 0 || k < 0) return 0;
        return aCb(n + k - 1, k); }}; // nHk = (n+k-1)Ck : n is separator
typedef ModInt<1000000007> mint;
typedef long double ld;
ld lginf = -1;
ll mul(ll a, ll b) {
    if (lginf < 0) lginf = log10l(infl); ld aa = log10l(a), bb = log10l(b);
    if (lginf <= aa + bb) return infl; return a * b;
}
ll gcd(ll a, ll b) { return a ? gcd(b%a, a) : b; }
ll lcm(ll a, ll b) { if (a == infl || b == infl) return infl; return mul(a / gcd(a, b), b); }
/*---------------------------------------------------------------------------------------------------
　　　　　　　　　　　 ∧＿∧  
　　　　　 ∧＿∧ 　（´<_｀ ）　 Welcome to My Coding Space!
　　　　 （ ´_ゝ`）　/　 ⌒i     
　　　　／　　　＼　 　  |　|     
　　　 /　　 /￣￣￣￣/　　|  
　 ＿_(__ﾆつ/　    ＿/ .| .|＿＿＿＿  
　 　　　＼/＿＿＿＿/　（u　⊃  
---------------------------------------------------------------------------------------------------*/









int N; ll M;
ll A[20]; double P[20];
double sm[1 << 20];
//---------------------------------------------------------------------------------------------------
void _main() {
    cin >> N >> M;
    rep(i, 0, N) cin >> A[i];
    rep(i, 0, N) cin >> P[i], P[i] = P[i] / 100;
    
    rep(msk, 1, 1 << N) {
        ll lc = 1;
        rep(i, 0, N) if (msk & (1 << i)) lc = lcm(lc, A[i]);
        sm[msk] = M / lc;
    }
    rep(i, 0, N) rep(j, 0, 1 << N) if (j & (1 << i)) sm[j] -= sm[j ^ (1 << i)];

    double ans = 0;
    rep(msk, 0, 1 << N) {
        double prop = 1;
        rep(i, 0, N) {
            if (msk & (1 << i)) prop *= P[i];
            else prop *= 1 - P[i];
        }
        ans += abs(sm[msk]) * prop;
    }
    printf("%.10f\n", ans);
}