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>).
y, n, s = 0, int(input()), list(map(int, input().split()))
for i in range(n - 1):
    t = s[i]
    if t ^ y >= t: continue
    k = s[i + 1]
    if k < t:
        x = k ^ t
        if x > y: y = x
        j = i + 2
        while j < n and s[j] < t:
            if s[j] > k:
                k = s[j]
                x = k ^ t
                if x > y: y = x
            j += 1
for i in range(1, n):
    t = s[i]
    if t ^ y >= t: continue
    k = s[i - 1]
    if k < t:
        x = k ^ t
        if x > y: y = x
        j = i - 2
        while j >= 0 and s[j] < t:
            if s[j] > k:
                k = s[j]
                x = k ^ t
                if x > y: y = x
            j -= 1
print(y)