#include <bits/stdc++.h>
using namespace std;
#include <atcoder/modint>
using namespace atcoder;
//using mint = modint998244353;
using mint = modint1000000007;
using ll = long long;
using ld = long double;
using pll = pair<ll, ll>;
using tlll = tuple<ll, ll, ll>;
constexpr ll INF = 1LL << 60;
template<class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
ll safemod(ll A, ll M) {return (A % M + M) % M;}
ll divfloor(ll A, ll B) {if (B < 0) {return divfloor(-A, -B);} return (A - safemod(A, B)) / B;}
ll divceil(ll A, ll B) {if (B < 0) {return divceil(-A, -B);} return divfloor(A + B - 1, B);}
template<class T> void unique(vector<T> &V) {V.erase(unique(V.begin(), V.end()), V.end());}
template<class T> void sortunique(vector<T> &V) {sort(V.begin(), V.end()); V.erase(unique(V.begin(), V.end()), V.end());}
#define FINALANS(A) do {cout << (A) << '\n'; exit(0);} while (false)
template<class T> void printvec(vector<T> &V) {int _n = V.size(); for (int i = 0; i < _n; i++) cout << V[i] << (i == _n - 1 ? "" : " ");cout << '\n';}
template<class T> void printvect(vector<T> &V) {for (auto v : V) cout << v << '\n';}
template<class T> void printvec2(vector<vector<T>> &V) {for (auto &v : V) printvec(v);}

int main()
{
  ll N;
  cin >> N;
  vector<ll> A(N), B(N);
  for (ll i = 0; i < N; i++)
  {
    cin >> A.at(i);
  }
  for (ll i = 0; i < N; i++)
  {
    cin >> B.at(i);
  }

  vector<ll> S(N);
  S.at(0) = B.at(0);
  for (ll i = 0; i < N - 1; i++)
  {
    S.at(i + 1) = S.at(i) ^ B.at(i + 1);
  }
  //printvec(S);

  const ll K = 32;
  vector<ll> cnt(K, 0);
  ll cnt0 = 0, cnt1 = 0;
  deque<ll> deq;
  ll ans = 0;
  for (ll i = 0; i < N; i++)
  {
    deq.push_back(i);
    for (ll k = 0; k < K; k++)
    {
      if (A.at(i) & (1LL << k))
        cnt.at(k)++;
    }

    while (*max_element(cnt.begin(), cnt.end()) >= 2)
    {
      ll j = deq.front();
      deq.pop_front();

      for (ll k = 0; k < K; k++)
      {
        if (A.at(j) & (1LL << k))
          cnt.at(k)--;
      }

      if (S.at(j) == 0)
        cnt0--;
      else
        cnt1--;
    }

    ll tmp = (S.at(i) == 0 ? cnt0 : cnt1) + (B.at(i) == 0 ? 1 : 0);
    ans += tmp;

    //cerr << i << "  " << deq.size() << "  " << cnt0 << " " << cnt1 << "  " << tmp << endl;

    if (S.at(i) == 0)
      cnt0++;
    else
      cnt1++;
  }
  cout << ans << endl;
}