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>
using namespace std;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,a,b) for(int i=(a);i<(b);i++)
#define pb push_back
#define eb emplace_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second

using vint=vector<int>;
using pint=pair<int,int>;
using vpint=vector<pint>;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

template<class A,class B>
ostream& operator<<(ostream& ost,const pair<A,B>&p){
	ost<<"{"<<p.first<<","<<p.second<<"}";
	return ost;
}

template<class T>
ostream& operator<<(ostream& ost,const vector<T>&v){
	ost<<"{";
	for(int i=0;i<v.size();i++){
		if(i)ost<<",";
		ost<<v[i];
	}
	ost<<"}";
	return ost;
}

int msb(int x){
	return 63-__builtin_clzll(x);
}

int N;
int A[111111];

signed main(){
	cin>>N;
	rep(i,N)cin>>A[i];

	int base=0;
	rep(i,N)base^=A[i];

	rep(i,60){
		if(!(base>>i&1))continue;
		rep(j,N)if(A[j]>>i&1)A[j]^=1ll<<i;
	}

	for(int i=0;i<N;i++){
		for(int j=i+1;j<N;j++){
			if(A[i]<A[j])swap(A[i],A[j]);
		}
		if(A[i]==0)break;
		for(int j=0;j<N;j++){
			if(i==j)continue;
			if(A[j]>>msb(A[i])&1)A[j]^=A[i];
		}
	}


	int X=0;
	rep(i,N)X^=A[i];
	int Y=base^X;
	cout<<base+(X&Y)*2<<endl;
	return 0;
}