281_D. Maximum Xor Secondary
Bike loves looking for the second maximum element in the sequence. The second maximum element in the sequence of distinct numbers x1, x2, ..., xk (k > 1) is such maximum element xj, that the following inequality holds: <image>.

The lucky number of the sequence of distinct positive integers x1, x2, ..., xk (k > 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 s1, s2, ..., sn (n > 1). Let's denote sequence sl, sl + 1, ..., sr as s[l..r] (1 ≤ l < 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.

Input

The first line contains integer n (1 < n ≤ 105). The second line contains n distinct integers s1, s2, ..., sn (1 ≤ si ≤ 109).

Output

Print a single integer — the maximum lucky number among all lucky numbers of sequences s[l..r].

Examples

Input

5
5 2 1 4 3


Output

7


Input

5
9 8 3 5 7


Output

15

Note

For the first sample you can choose s[4..5] = {4, 3} and its lucky number is (4 xor 3) = 7. You can also choose s[1..2].

For the second sample you must choose s[2..5] = {8, 3, 5, 7}.
```python
#https://codeforces.com/problemset/problem/281/D

def empty(l):
    return len(l) == 0

def lucky_number(arr):
    st = []
    l = 0

    for d in arr:
        while not empty(st) and st[-1] < d:
            l = max(l, st.pop() ^ d)
        if not empty(st):
            l = max(l, st[-1] ^ d)
        st.append(d)
        # print(st)

    # print(l)
    # print(st)


    d = 0
    if not empty(st):
        d = st.pop()
    while not empty(st):
        l = max(l, d ^ st[-1])
        d = st.pop()
    return l

N = int(input())
arr = list(map(int, input().split()))
print(lucky_number(arr))

# for i in range(len(arr)):
#     for j in range(i + 1, len(arr)):
#         print(arr[i], arr[j], arr[i]^arr[j])
```