#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#define pli pair<ll, int>
#define pil pair<int, ll>
#define pll pair<ll, ll>
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep3(i, l, n) for (int i = l; i < (n); ++i)
#define sz(v) (int)v.size()
#define endl '\n'
const int inf = 1000000007;
const ll INF = 1e18;
// int mod = 998244353;
int mod = 1000000007;
#define abs(x) (x >= 0 ? x : -(x))
#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())
#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
template<typename T, typename U> T pow_(T a, U b) { return b ? pow_(a * a, b / 2) * (b % 2 ? a : 1) : 1; }
ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }
template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } return os; }
template<typename T> inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }
template<class T, class T2> inline void add(T &a, T2 b) { a += b % mod; if (a >= mod) a -= mod; }
template<class T> void operator += (vector<T>& v, vector<T> v2) { rep(i, sz(v2)) v.eb(v2[i]); }

void solve();

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(12); // debuged
    int T;
    T = 1;
    while (T--) solve();
}

// https://betrue12.hateblo.jp/entry/2019/11/22/012255
// 行基本変形によって, xor和のmaxが変わらない
// 赤=青(1の個数が奇数個の桁を除いた)だから, 赤だけ考える
// Aのうちいくつかを選ぶ
// そのありうる集合が, 行基本変形によって変わらない
void solve() {
    int n;
    cin >> n;
    vector<ll> a(n);
    cin >> a;
    auto out = [&]() { cerr << endl; rep(i, n) { rep(j, 61) cerr << !!(a[i] & (1LL << (60 - j))) << " "; cerr << endl; } };
    
    ll ans = 0;
    rep(i, 61) { // 1が奇数個の桁は別
        int cnt = 0;
        rep(j, n) if (a[j] & (1LL << i)) ++cnt;
        if (cnt % 2) {
            ans += 1LL << i;
            rep(j, n) if (a[j] & (1LL << i)) a[j] -= 1LL << i;
        }
    }
    // out();
    
    int done = 0;
    rep(i, 61) { // 掃き出し法
        int k = -1;
        rep3(j, done, n) { // 列doneを1にする行
            if (a[j] & (1LL << (60 - i))) {
                k = j;
                break;
            }
        }
        if (k == -1) continue;
        // cerr << "i " << i << endl;
        swap(a[done], a[k]);
        // 他の行とxor
        rep(j, n) if (j != done && (a[j] & (1LL << (60 - i)))) a[j] ^= a[done];
        ++done;
    }
    // out();
    
    rep(i, 61) {
        int cnt = 0;
        rep(j, n) if (a[j] & (1LL << i)) ++cnt;
        if (cnt % 2) ans += 2LL << i;
    }
    cout << ans << endl;
}


Measured memory_bytes: 3972