Bike loves looking for the second maximum element in the sequence. The second maximum element in the sequence of distinct numbers *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ..., *x*<sub class="lower-index">*k*</sub> (*k* &gt; 1) is such maximum element *x*<sub class="lower-index">*j*</sub>, that the following inequality holds: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/41e9a774451aa769280692592b29103e0a6ecf97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

The lucky number of the sequence of distinct positive integers *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ..., *x*<sub class="lower-index">*k*</sub> (*k* &gt; 1) is the number that is equal to the bitwise excluding OR of the maximum element of the sequence and the second maximum element of the sequence.

You've got a sequence of distinct positive integers *s*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub>, ..., *s*<sub class="lower-index">*n*</sub> (*n* &gt; 1). Let's denote sequence *s*<sub class="lower-index">*l*</sub>, *s*<sub class="lower-index">*l* + 1</sub>, ..., *s*<sub class="lower-index">*r*</sub> as *s*[*l*..*r*] (1 ≤ *l* &lt; *r* ≤ *n*). Your task is to find the maximum number among all lucky numbers of sequences *s*[*l*..*r*].

Note that as all numbers in sequence *s* are distinct, all the given definitions make sence.
The first line contains integer *n* (1 &lt; *n* ≤ 10<sup class="upper-index">5</sup>). The second line contains *n* distinct integers *s*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub>, ..., *s*<sub class="lower-index">*n*</sub> (1 ≤ *s*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>).
import sys 
input = sys.stdin.buffer.readline 

def process(A):
    n = len(A)
    R = 2**60 
    while True:
        L = []
        for i in range(n):
            if A[i]&R > 0:
                L.append(i)
        if len(L)==n:
            for i in range(n):
                A[i]-=R 
        elif len(L) < n:
            break 
        R = R//2 
    answer = 0
    for i1 in L:
        my_max = 0 
        for j in range(i1+1, n):
            if R&A[j] > 0:
                break 
            my_max = max(my_max, A[j])
            answer = max(answer, my_max^A[i1])
        my_max = 0 
        for j in range(i1-1, -1, -1):
            if R&A[j] > 0:
                break 
            my_max = max(my_max, A[j])
            answer = max(answer, my_max^A[i1])
    sys.stdout.write(f'{answer}\n')
    
    
n = int(input())
A = [int(x) for x in input().split()]
process(A)
            
    