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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

const SZ:usize = 60;
fn main() {
    input! {
        n:usize,
        mut a:[i64;n]
    };

    a.sort();
    let a:Vec<_> = a.into_iter().rev().collect();
    let mut bases = vec![0;SZ];

    let sum = a.iter().fold(0,|acc,v|acc^v);
    let flag:Vec<_> = (0..SZ).map(|i| (sum>>i)&1==1).collect();

    let tpb = |x| {
        for i in (0..SZ).rev() {
            if flag[i] { continue; }
            if ((x>>i)&1)==1 {
                return i;
            }
        }
        return SZ;
    };
    for &v in &a {
        let mut x = v;
        for i in (0..SZ).rev() {
            if flag[i] { continue; }
            if ((x>>i)&1)==1 && bases[i]!=0 {
                x^=bases[i];
            }
        }
        if x!=0 {
            let b = tpb(x);
            if b<SZ {
                bases[b] = x;
            }
        }
    }
    // eprintln!("{:?}",bases);
    
    let mut av = 0;
    for i in (0..SZ).rev() {
        if (av>>i)&1==0 {
            av^=bases[i];
        }
    }
    let bv = sum^av;
    // eprintln!("a {} b {}",av,bv);
    println!("{}",av+bv);

}