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 ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#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 srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define svec(v) cout<<#v<<":";rep(kbrni,v.size())cout<<" "<<v[kbrni];cout<<endl
#define sset(s) cout<<#s<<":";each(kbrni,s)cout<<" "<<kbrni;cout<<endl
#define smap(m) cout<<#m<<":";each(kbrni,m)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 100005;

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

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    ll m;
    cin >> n >> m;
    vl a(n);
    vd p(n);
    rep(i,n){
        cin >> a[i];
    }
    rep(i,n){
        cin >> p[i];
    }
    double two[21];
    two[0] = 1.0;
    rep(i,20){
        two[i+1] = two[i] * 2;
    }
    double ans = 0;
    srep(i,1,(1 << n)){
        double res = 1.0;
        ll val = 1;
        bool flag = true;
        rep(j,n){
            if((i >> j) & 1){
                res *= p[j]/100.0;
                if(flag){
                    ll g = gcd(val,a[j]);
                    double ch = val;
                    if(ch/g*a[j] > m+1000000LL){
                        flag = false;
                        break;
                    }else{
                        val = val/g*a[j];
                    }
                }
            }
        }
        if(flag){
            // show(val);
            int num = __builtin_popcount(i);
            res *= m/val;
            // show(res);
            // res *= two[n-num];
            if(num % 2){
                ans += res;
            }else{
                ans -= res;
            }
        }
    }
    printf("%.12f\n",ans);
    return 0;
}