#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
#define forn(i, a, n) for (int i = (int)(a); i < (int)(n); ++i)
#define ford(i, a, n) for (int i = (int)(n) - 1; i >= (int)(a); --i)
#define fore(i, a, n) for (int i = (int)(a); i <= (int)(n); ++i)
#define all(a) (a).begin(), (a).end()
#define fs first
#define sn second
#define trace(a)\
for (auto i : a) cerr << i << ' ';\
cerr << '\n'
#define eb emplace_back
#ifndef M_PI
const ld M_PI = acos(-1.0);
#endif
template<typename T>
inline void setmax(T& a, T b) {
if (a < b) a = b;
}
template<typename T>
inline void setmin(T& a, T b) {
if (a > b) a = b;
}
template<typename T, typename S>
istream& operator>> (istream& in, pair<S, T>& p) {
in >> p.fs >> p.sn;
return in;
}
template<typename T, typename S>
ostream& operator<< (ostream& out, pair<S, T>& p) {
out << p.fs << ' ' << p.sn << ' ';
return out;
}
template<typename T>
istream& operator>> (istream& in, vector<T>& v) {
for (T& x : v) in >> x;
return in;
}
template<typename T>
ostream& operator<< (ostream& out, vector<T>& v) {
for (T& x : v) out << x << ' ';
return out;
}
const ld eps = 1e-9;
const int INF = 2000000000;
const ll LINF = 1ll * INF * INF;
const ll MOD = 1000000007;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
srand((unsigned)chrono::high_resolution_clock::now().time_since_epoch().count());
int n;
cin >> n;
vector<int> a(n), b(n);
int ans = 0;
cin >> a >> b;
forn(i, 0, 30) {
//cerr << i << '\n';
vector<int> res(n);
forn(j, 0, n) res[j] = b[j] % (1 << (i + 1));
sort(all(res));
int total = 0;
forn(j, 0, n) {
int x = a[j] % (1 << (i + 1));
int st = x ? (1 << (i + 1)) - x : 0;
int fn = st ^ (1 << i);
//cerr << x << ' ' << st << ' ' << fn << '\n';
int cnt = (int)(lower_bound(all(res), st) - res.begin()) - (int)(lower_bound(all(res), fn) - res.begin());
if (cnt < 0) cnt += n;
total += cnt;
}
//cerr << total << '\n';
if (total % 2)
ans ^= (1 << i);
}
cout << ans << '\n';
}