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 <bits/stdc++.h>
using namespace std;
#define rep(i, n)  for(long long i=0;i<(long long)(n);i++)
#define REP(i,k,n) for(long long i=k;i<(long long)(n);i++)
#define all(a) a.begin(),a.end()
#define pb push_back
#define eb emplace_back
#define lb(v,k) (lower_bound(all(v),k)-v.begin())
#define ub(v,k) (upper_bound(all(v),k)-v.begin())
#define fi first
#define se second
typedef long long ll;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> PP;
typedef multiset<ll> S;
typedef priority_queue<ll> PQ;
typedef priority_queue<P,vector<P>,greater<P>> SPQ;
using vi=vector<ll>;
using vvi=vector<vi>;
using vvvi=vector<vvi>;
const ll inf=1001001001001001;
const int INF=1001001001;
const int mod=1000000007;
const double pi=3.14159265358979323846;
bool chmin(auto &a,auto b){if(a>b){a=b;return true;}return false;}
bool chmax(auto &a,auto b){if(a<b){a=b;return true;}return false;}
void outvi(vi v){rep(i,v.size()){if(i)cout<<' ';cout<<v[i];}cout<<endl;}
void outvvi(vvi v){rep(i,v.size()){rep(j,v[i].size()){if(j)cout<<' ';cout<<v[i][j];}cout<<endl;}}
int main(){
    ll n;cin>>n;
    vi v(n);
    ll sum=0,ans=0;
    rep(i,n)cin>>v[i];
    rep(i,n)sum^=v[i];
    vector<bool> used(n,false);
    for(ll i=62;i>=0;i--){
        if(sum>>i&1){
            ans+=((ll)(1))<<i;
            continue;
        }
        ll w=inf;
        rep(j,n)if(!used[j]&&(v[j]>>i&1))w=j;
        if(w==inf){
            ll t=0;
            rep(j,n)if(v[j]>>i&1)t^=1;
            if(t)ans+=((ll)(1))<<i+1;
            continue;
        }
        used[w]=true;
        rep(j,n){
            if(j==w)continue;
            if(v[j]>>i&1)v[j]^=v[w];
        }
        ans+=((ll)(1))<<(i+1);
        cout<<ans<<endl;
    }
    cout<<ans<<endl;
}