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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {

    public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      int num = Integer.parseInt(sc.next());
      
      long[] inputs = new long[num];
      long sum = 0;
      for (int i = 0; i < num; i++) {
        inputs[i] = Long.parseLong(sc.next());
        sum ^= inputs[i];
      }
      // only interested in 1's bit
      for (int i = 0; i < num; i++) {
        inputs[i] &= ~sum;
      }
      System.out.println(sum + (maxSubarrayXOR(inputs) << 1));
    }
  
  	public static long maxSubarrayXOR(long A[]) {
      int idx = 0;
      for (int bit = 60; bit >= 0; bit--) {
        int find = idx;

        while (find < A.length) {
          // find the element which MSB is in bit'th
          if ((A[find] & (1L << bit)) != 0) {
          	break;
          }
          find++;
        }
        
        if (find == A.length) continue;
        
        // swap element in idx and find
        if (idx != find) {
          A[idx] ^= A[find];
          A[find] ^= A[idx];
          A[idx] ^= A[find];
          find = idx;
        }
        
        for (int i = 0; i < A.length; i++) {
       	  if (i != find && (A[i] & (1L << bit)) != 0) {
            A[i] ^= A[find];
          }
        }

        idx++;
      }
      
      long ret = 0;
      for (long l: A) {
        ret ^= l;
      }

      return ret;
    } 
}