Problem:
Score: 600 points

Problem Statement

We have N non-negative integers: A_1, A_2, ..., A_N.

Consider painting at least one and at most N-1 integers among them in red, and painting the rest in blue.

Let the beauty of the painting be the \mbox{XOR} of the integers painted in red, plus the \mbox{XOR} of the integers painted in blue.

Find the maximum possible beauty of the painting.

What is \mbox{XOR}?

The bitwise \mbox{XOR} x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non-negative integers x_1, x_2, \ldots, x_n is defined as follows:

When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even.

For example, 3 \oplus 5 = 6.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

0 \leq A_i < 2^{60}\ (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible beauty of the painting.

Sample Input 1

3
3 6 5

Sample Output 1

12

If we paint 3, 6, 5 in blue, red, blue, respectively, the beauty will be (6) + (3 \oplus 5) = 12.

There is no way to paint the integers resulting in greater beauty than 12, so the answer is 12.

Sample Input 2

4
23 36 66 65

Sample Output 2

188

Sample Input 3

20
1008288677408720767 539403903321871999 1044301017184589821 215886900497862655 504277496111605629 972104334925272829 792625803473366909 972333547668684797 467386965442856573 755861732751878143 1151846447448561405 467257771752201853 683930041385277311 432010719984459389 319104378117934975 611451291444233983 647509226592964607 251832107792119421 827811265410084479 864032478037725181

Sample Output 3

2012721721873704572

A_i and the answer may not fit into a 32-bit integer type.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define GO ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
const int N = 16;
const int M = 1e3 + 5;
const int Mod = 998244353;

struct Base
{
    int m;
    vector <ll> A;

    Base(int m = 0) : m(m)
    {
        A.assign(max(m, 1), 0);
    }

    void insert(ll x)
    {
        for (ll i = m - 1; ~i; --i)
        {
            if (x & (1LL << i))
            {
                if (A[i])
                {
                    x ^= A[i];
                }

                else
                {
                    A[i] = x;
                    break;
                }
            }
        }
    }

    vector <ll> Get()
    {
        vector <ll> Res;

        for (int i = 0; i < m; i++)
            if (A[i])
                Res.push_back(A[i]);

        return Res;
    }
};

int main()
{
    GO;
    int n;
    cin >> n;
    ll XOR = 0;
    vec a(n);
    for (auto& i : a)
        cin >> i, XOR ^= i;
    ll sum = XOR;
    for (auto& i : a)
        i &= ~XOR;
    Base b(60);
    for (auto i : a)
        b.insert(i);
    XOR = (1LL << 60) - 1 - XOR;
    ll temp = XOR;
    for(int i=59;i>=0;i--)
        if (XOR & 1LL << i)
        {
            if (temp & 1LL << i)
                temp ^= b.A[i];
            if (~temp & 1LL << i)
                sum += 2LL << i;
        }
    cout << sum<< endl;
    return 0;
}