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 heapq;
import math;
from collections import deque, Counter;
import sys;

mod = 998244353;
input = sys.stdin.readline;







n = int(input())


array = list(map(int, input().split()))



ans = 0;



def check(currNum):
    numOne, numTwo = None, None;


    for num in array:
        if (currNum & num) == currNum:
            if numTwo is not None:
                return True;
            numOne = num;
        elif not (currNum & num):
            if numOne is not None:
                return True;
            numTwo = num;
        else:
            if numOne is not None and num > numOne:
                numOne = None;

            if numTwo is not None and num > numTwo:
                numTwo = None;

    return False;


for i in range(32, -1, -1):
    if check((1 << i)):
        ans |= (1 << i)
        for i in range(len(array)):
            if array[i] & (1 << i):
                array[i] ^= (1 << i)

print(ans)