//g++ 2.cpp -std=c++14 -O2 -I .
#include <bits/stdc++.h>
using namespace std;

#include <atcoder/all>
using namespace atcoder;

using ll = long long;
using ld = long double;

using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vld = vector<ld>;
using vvld = vector<vld>;
using vst = vector<string>;
using vvst = vector<vst>;

#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define pq_big(T) priority_queue<T,vector<T>,less<T>>
#define pq_small(T) priority_queue<T,vector<T>,greater<T>>
#define all(a) a.begin(),a.end()
#define rep(i,start,end) for(ll i=start;i<(ll)(end);i++)
#define per(i,start,end) for(ll i=start;i>=(ll)(end);i--)
#define uniq(a) sort(all(a));a.erase(unique(all(a)),a.end())

int main(){
  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  /*
  B xor = 0
  A bit互いに素 32個以上あると詰み

  B xor = 1
  A bit1だけ共通 

  100個ぐらい見れば十分かな
  */

  int n;
  cin>>n;

  ll ans=0;
  vi a(n);
  vi b(n);
  rep(i,0,n)cin>>a[i];
  rep(i,0,n)cin>>b[i];

  rep(i,0,n){
    ll asum=0,axor=0,bxor=0;
    rep(j,0,100){
      if(i+j>=n)break;
      asum+=a[i+j];
      axor^=a[i+j];
      bxor^=b[i+j];

      if(asum-axor==bxor)ans++;
    }
  }

  cout<<ans<<endl;
}