// kinen submit
#include <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <sstream>
#include <functional>
#include <map>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <list>
#include <numeric>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
typedef pair<P,ll> PP;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const ll INF = 1LL<<29;
const ll mod = 1e9+7;
#define rep(i,n) for(int (i)=0;(i)<(ll)(n);++(i))
#define repd(i,n,d) for(ll (i)=0;(i)<(ll)(n);(i)+=(d))
#define all(v) (v).begin(), (v).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset((m),(v),sizeof(m))
#define chmin(X,Y) ((X)>(Y)?X=(Y),true:false)
#define chmax(X,Y) ((X)<(Y)?X=(Y),true:false)
#define fst first
#define snd second
#define UNIQUE(x) (x).erase(unique(all(x)),(x).end())
template<class T> ostream &operator<<(ostream &os, const vector<T> &v){int n=v.size();rep(i,n)os<<v[i]<<(i==n-1?"":" ");return os;}

#define N 100100
ll n;
ll a[2][N], c[2][N], p[2][N], pr[2][N], used[N], ix[N], lp[N], dd[N], rr[N];
map<PP, vector<ll>> s;

void extgcd(ll x, ll y, ll &a, ll &b, ll &g){
	if(y == 0){
		a = 1;
		b = 0;
		g = x;
	} else {
		extgcd(y, x%y, a, b, g);
		swap(a, b);
		b -= a*x/y;
	}
}

int main(){
	scanf("%d", &n);
	rep(i, 2) rep(j, n){
		scanf("%lld", a[i]+j);
		a[i][j]--;
	}
	mset(p, -1);
	rep(i, 2) rep(j, n){
		if(p[i][j] != -1) continue;
		ll k = a[i][j];
		p[i][j] = 1;
		while(k != j){
			c[i][k] = p[i][j]++;
			pr[i][k] = j;
			k = a[i][k];
		}
		pr[i][j] = j;
	}
	ll res = 0;
	rep(j, n){
		ll b[2], d;
		extgcd(p[0][j], p[1][j], b[0], b[1], d);
		ll r[2], q[2];
		rep(i, 2) r[i] = c[i][j]%d;
		if(r[1]<r[0]) r[1] += d;
		rep(i, 2) q[i] = c[i][j]/d;
		ll qq = q[1]-q[0];
		b[0] *= qq; b[0] %= p[1][j]/d; if(b[0]<0) b[0] += p[1][j]/d;
		b[1] * -qq; b[1] %= p[0][j]/d; if(b[1]<0) b[1] += p[0][j]/d;
		ix[j] = b[0]*p[0][j]*d+r[0];
		lp[j] = p[0][j]/d*p[1][j]/d;
		dd[j] = d;
		rr[j] = r[1]-r[0];
		s[PP(P(pr[0][j], pr[1][j]), r[j])].push_back(j);
	}
	for(auto &&x: s){
		auto &&v = x.snd;
		int m = v.size();
		vector<ll> vv(m+1);
		rep(i, m) vv[i] = ix[v[i]];
		v[m] = vv[0]+lp[v[0]];
		rep(i, m){
			ll y = vv[i+1]-vv[i]-1;
			res += y*(y+1)/2;
			res %= mod;
		}
	}
	cout<<res<<endl;
	return 0;
}