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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <vector>

using namespace std;

#define l first
#define r second
#define int long long

int n;
long long a[500000];
vector<long long> multi;

signed main()
{
    cin >> n;
    long long x = 0;
    for (int i = 0; i < n; i++) cin >> a[i], x ^= a[i];
    for (int i = 0; i < n; i++) a[i] -= a[i] & x;
    for (int i = 0; i < n; i++)
    {
        int kek = 0;
        long long bita = 1;
        for (int j = 0; j < 60; j++)
        {
            if ((1LL << j) & a[i]) bita = (1LL << j);
        }
        for (int j = 0; j < multi.size(); j++)
        {
            if ((multi[j] & a[i]) >= bita) kek = 1;
        }
        if (kek)
        {
            long long cur = a[i];
            for (int i = 0; i < multi.size(); i++)
            {
                if (cur & multi[i] >= bita)
                {
                    cur ^= multi[i];
                    if (cur == 0) break;
                    while (bita >= 1 && !(cur & bita))
                    {
                        bita /= 2;
                    }
                }
            }
            if (cur != 0 && bita == 0)
            {
                multi.push_back(cur);
                sort(multi.begin(), multi.end());
                reverse(multi.begin(), multi.end());
            }
        }
        else
        {
            multi.push_back(a[i]);
            sort(multi.begin(), multi.end());
            reverse(multi.begin(), multi.end());
        }
    }
    sort(multi.begin(), multi.end());
    reverse(multi.begin(), multi.end());
    long long res = 0;
    long long b = (1LL << 61);
    for (int i = 0; i < multi.size(); i++)
    {
        //cout << multi[i] << "\n";
        while (b > multi[i]) b /= 2;
        if ((res & b) == 0) res ^= multi[i];
    }
    cout << x + 2 * res;
}