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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.math.*

fun main(args: Array<String>) {
    val N = readInputLine().toInt()
    var aList = readInputLine().split(" ").map { it.toLong() }
    
    val allXor = aList.fold(0L, { all, elem -> all xor elem })
    val aArr = aList.map { it and allXor.inv() }.toLongArray()
    
    var rank = 0
    for (i in 59 downTo 0) {
        var j = rank
        while (j < N) {
            if (aArr[j] and (1L shl i) != 0L) {
                break
            }
            j++
        }
        if (j == N) {
            continue
        }
        
        if (j > rank) {
            aArr[rank] = aArr[rank] xor aArr[j]
        }
        
        for (j in rank + 1 until N) {
            aArr[j] = Math.min(aArr[j], aArr[j] xor aArr[rank])
        }
        
        rank++
    }
    
    println(aArr.fold(0L) { acc, a ->
        Math.max(acc, acc xor a)
    } * 2L + allXor)
}

fun readInputLine(): String {
    return readLine()!!
}