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>).
n = int(input())
a = list(map(int, input().split()))

stack = []
res = 0
for i, v in enumerate(a):
    while stack and stack[-1] < v:
        res = max(res, stack.pop() ^ v)
    if stack:
        res = max(res, stack[-1] ^ v)
    stack.append(v)

print(res)