#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double db;
const int N = 5e5+10;
int n,p[N],q[N];  db ans;
int dfn[N],ord[N],sz[N],dep[N],dtot;
int rt[N]; vector<int> e[N];

void dfs(int u){
    dfn[u] = ++dtot,ord[dtot] = u,sz[u] = 1;
    for(auto v : e[u]){
        dep[v] = dep[u] + 1;
        dfs(v);
        sz[u] += sz[v];
    }
}
db w[N],vt[N],k[N],y[N];

void calc(int x){
    int rt = ::rt[x],L = dfn[rt],R = dfn[rt] + sz[rt] - 1; 
    w[rt] = 0; for(int i=L+1;i<=R;i++)w[ord[i]] = 1.00 * ((q[ord[i]]+x)%n) / n;
    
    for(int i=L+1;i<=R;i++){
        int u = ord[i];
        vt[u] = sz[u] / (1-w[u]); //xu = vt[u] * (yu - yfa)
    }
    for(int i=R;i>L;i--){
        int u = ord[i];
        db KL = 0,KR = 0; //yu 的系数 和 yfa 的系数
        KL += vt[u],KR += vt[u];
        KL += -1;
        for(auto v : e[u]){
            KL -= w[v]*vt[v]*(k[v]-1);
        }
        k[u] = KR / KL;
    }
    //回代求出每个点是y_rt的多少倍
    //y_rt = 1/sz_rt * x_rt，借此解出y[rt]，再回代解出y
    k[rt] = 1; db sum = sz[rt];
    for(int i=L+1;i<=R;i++){
        int u = ord[i];
        k[u] *= k[p[u]];
        sum += vt[u] * (k[u] - k[p[u]]);
    }
    //sum * y_rt = 1
    y[rt] = 1/sum;

    db res = 0;
    for(int i=L+1;i<=R;i++){
        int u = ord[i];
        y[u] = k[u] * y[rt];

        db X = vt[u] * (y[u] - y[p[u]]);
        res += X * w[u];
    } 
    ans = max(ans,res);
}

int main(){
    scanf("%d",&n); for(int i=1;i<=n;i++)scanf("%d",&p[i]),e[p[i]].push_back(i);
    for(int i=1;i<=n;i++)cin>>q[i],rt[(n-q[i])%n] = i;
    dfs(1);
    
    for(int x=0;x<n;x++)calc(x);

    printf("%.9Lf\n",ans);
    return 0;
}