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, n) for (int i = 0; i < int(n); i++)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; i--)
#define reps(i, n) for (int i = 1; i <= int(n); i++)
#define rreps(i, n) for (int i = int(n); i >= 1; i--)
#define repc(i, n) for (int i = 0; i <= int(n); i++)
#define rrepc(i, n) for (int i = int(n); i >= 0; i--)
#define repi(i, a, b) for (int i = int(a); i < int(b); i++)
#define repic(i, a, b) for (int i = int(a); i <= int(b); i++)
#define all(a) (a).begin(), (a).end()
#define bit32(x) (1 << (x))
#define bit64(x) (1ll << (x))
#define sz(v) ((int) v.size())

using namespace std;

using i64 = long long;
using u64 = unsigned long long;
using f80 = long double;
using vi32 = vector<int>;
using vi64 = vector<i64>;
using vf80 = vector<f80>;
using vstr = vector<string>;

void yes() { cout << "Yes" << endl; exit(0); }
void no() { cout << "No" << endl; exit(0); }
template <typename T> class pqasc : public priority_queue<T, vector<T>, greater<T>> {};
template <typename T> class pqdesc : public priority_queue<T, vector<T>, less<T>> {};
template <typename T> void amax(T &x, T y) { x = max(x, y); }
template <typename T> void amin(T &x, T y) { x = min(x, y); }
template <typename T> T exp(T x, i64 n, T e = 1) { T r = e; while (n > 0) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
template <typename T> istream& operator>>(istream &is, vector<T> &v) { for (auto &x : v) is >> x; return is; }
template <typename T> ostream& operator<<(ostream &os, vector<T> &v) { rep(i, v.size()) { if (i) os << ' '; os << v[i]; } return os; }
void solve(); int main() { ios::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(16); solve(); return 0; }

const int INF = 1001001001;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};

u64 gcd(u64 a, u64 b) {
  if (min(a, b) == 0) {
    return max(a, b);
  }
  if (a % b == 0) {
    return b;
  } else {
    return gcd(b, a % b);
  }
}

u64 lcm(u64 a, u64 b) {
  if (min(a, b) == 0) {
    return max(a, b);
  }
  return a / gcd(a, b) * b;
}

void solve() {
  int n;
  u64 m;
  cin >> n >> m;
  vector<u64> a(n);
  vector<f80> p(n);
  cin >> a >> p;
  rep(i, n) p[i] /= 100;
  vector<f80> c(1 << n);
  auto check = [](u64 a, u64 b, u64 m) {
    u64 g = gcd(a, b);
    a /= g;
    int c = 0;
    while (a >= 10) c++, a /= 10;
    while (b >= 10) c++, b /= 10;
    if (c >= 19) return false;
    return a * b <= m;
  };
  reps(i, (1 << n) - 1) {
    u64 acc = 0;
    f80 percent = 1;
    bool ok = true;
    rep(j, n) if (i >> j & 1) {
      if (!check(acc, a[j], m)) {
        ok = false;
        break;
      }
      acc = lcm(acc, a[j]);
      percent *= p[j];
    }
    if (ok) {
      c[i] = (m / acc) * percent;
    }
  }
  rep(i, n) rep(s, 1 << n) if (s >> i & 1) {
    c[s] -= c[s ^ (1 << i)];
  }
  cout << abs(c[(1 << n) - 1]) << endl;
}