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: Runtime Error)

#include <bits/stdc++.h>
#include <cstdio>
#include <vector>
#include <iostream>
#include <utility>

#define fir first
#define sec second
#define get(n) scanf("%d",&n);
#define gets(s) string s;cin >> (s);
#define All(s) (c).begin(), (c).end()
#define chmin(x, y) x = min(x, y);
#define chmax(x, y) x = max(x, y);
#define rep(i, j) for (int (i)=0;(i)<(j);(i)++)
#define repk(i, j, k) for(int (i)=(j);(i)<(k);(i)++)
#define dump(x)  cout << #x << " = " << (x) << endl;
#define debug(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using namespace std;
typedef long long int ll;
const ll INF = 1e10;

int N;
vector<ll> num;

void input() {
    cin >> N;
    rep(i, N) {
        ll a;cin >> a;
        num.push_back(a);
    }
}

void solve() {
    ll a = num[0];
    for(int i=0;i<N;i++){
        a = a ^ num[i];
    }
    vector<int> b;//立っているbitを考える
    for (int i=60;i>0;i--){
        if((a & 1 << i)== 0){//全体の論理和のi桁にビットが立っていないなら
            b.push_back(i);
        }
    }
    vector<ll> bits;
    for (int i=0;i<N;i++){
        if((num[i] & 1 << b[0]))bits.push_back(num[i]);
    }
    
    for (int i=0;i<b.size();i++){
        for (int j=1;j<bits.size();i++){
            if(bits[j] != INF){
                if(bits[j] & 1 << b[i] == 0) bits[j] = INF;
            }
        }
    }
    ll ans = 0;
    for (int i=0;i<bits.size();i++){
        if( bits[i] != INF ){
            ans ^= bits[i];
        }
    }
    ll ans2 = a ^ ans ;
    ans += ans2;
    cout << ans << endl;
}

int main () {
    solve();
    return 0;
}