280_B. 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
#http://codeforces.com/problemset/problem/281/D

#SC = TC = o(n)

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

stack = []	#decreasing stack

max_xor = 0

for k in range(n):
	#if stack empty
	if not stack:
		stack.append(a[k])
	#if less than top of stack
	elif a[k]<stack[-1]:
		max_xor = max(max_xor, a[k]^stack[-1])
		stack.append(a[k])
	#input will be distinct so no equal condition
	#if greater than top of stack
	elif a[k]>stack[-1]:
		#pop and xor
		while stack and a[k]>stack[-1]:
			popped = stack.pop()
			max_xor = max(max_xor, a[k]^popped)
		#now if stack is not empty means below element is greater
		#so xor will it and push	#case 9 8 3 5 7
		if stack:
			max_xor = max(max_xor, a[k]^stack[-1])
		stack.append(a[k])

print(max_xor)
```