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}.
'''

10
4547989 39261040 94929326 38131456 26174500 7152864 71295827 77784626 89898294 68006331

'''



def maxxor(a,b,ans):
	t = a^b
	#print('{} {}'.format(a,b))
	if t > ans: return t
	else: return ans

n = int(input().strip())
arr = list(map(int, input().strip().split()))

s = list()
s.append(arr[0])
s.append(arr[1])
ans = maxxor(arr[0], arr[1], 0)
index = 2
while index < n:
	#print('ans = {} index = {} s = {}'.format(ans,index, s))
	if arr[index] > s[-1]:
		while (arr[index] > s[-1]): 
			t = s.pop()
			ans = maxxor(arr[index], t, ans)
			if  len(s) == 0: break
			
		
	if len(s) != 0:	ans = maxxor(arr[index], s[-1], ans)
	s.append(arr[index])
	
	index += 1
	
	
print(ans)
		
		
		
