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}.
l = int(input())
arr = [int(i) for i in input().split()]

stack = [arr[0]]
maxa = 0
x = 0

for i in range(1,len(arr)):
    # if arr[i] < stack[-1]:
    #     x = arr[i] ^ stack[-1]
    #     maxa = max(maxa,x)
    #     stack.append(arr[i])
    # else :
    while len(stack)!= 0 and arr[i] > stack[-1]:
        x = arr[i] ^ stack[-1]
        maxa = max(maxa,x)
        stack.pop()
    if len(stack) != 0:
        x = arr[i] ^ stack[-1]
        maxa = max(maxa,x)
    stack.append(arr[i])
    
print(maxa)
        