枚举中点，然后预处理出前缀最大和后缀最大，然后取个max就好

#include <bits/stdc++.h>
using namespace std;
long long R(){
	long long x = 0, f = 1;
	char ch = getchar();
	while(!isdigit(ch)){
		if(ch == '-') f = -1;
		ch = getchar();
	}
	while(isdigit(ch)){
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * f;
}

const long long N = 4e5 + 10;

long long n;

long long a[N];

long long mx[N], t[N*30][2], cnt;

void read(){
	n = R();
	for(long long i = 1;i <= n; i++) a[i] = R();
}

void insert(long long x){
	long long u = 0;
	for(long long i = 32;i >= 0; i--){
		long long w = ((x >> i) & 1);
		if(!t[u][w]) t[u][w] = ++cnt;
		u = t[u][w];
	}
	return ;
}

long long qry(long long x){
	long long u = 0;
	long long ans = 0;
	for(long long i = 32;i >= 0; i--){
		long long w = ((x >> i) & 1);
		if(t[u][(w^1)]){
			ans += (1<<i);
			u = t[u][(w^1)];
		}
		else{
			u = t[u][w];
		}
	}
	return ans;
}

void clear(){
	for(long long i = 0;i <= cnt; i++){
		t[i][1] = t[i][0] = 0;
	}
	cnt = 0;
}

void compute(){
	long long s = 0, l = 0, r = 0; 
	for(long long i = 0;i <= n; i++){
		s = (s ^ a[i]);
		insert(s);
		l = max(l,qry(s)) ;
		mx[i] = l;
	}
	clear();
	s = 0;
	long long ans = 0;
	for(long long i = n + 1;i >= 1; i--){
		s = (s ^ a[i]);
		insert(s);
		r = max(r,qry(s));
		ans = max(ans,r+mx[i-1]);
	}
	cout << ans;
}
int main(){
//	freopen("最大和ex.in","r",stdin);
//	freopen("最大和ex.out","w",stdout);
	read();
	compute();
	return 0;
}