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>).
# Python3 implementation of the approach

# Function to return the maximum possible xor
def maximumXor(arr: list, n: int) -> int:
	sForward, sBackward = [], []

	# To store the final answer
	ans = -1

	for i in range(n):

		# forward traversal
		while len(sForward) > 0 and arr[i] < arr[sForward[-1]]:
			ans = max(ans, arr[i] ^ arr[sForward[-1]])
			sForward.pop()

		sForward.append(i)

		# Backward traversal
		while len(sBackward) > 0 and arr[n - i - 1] < arr[sBackward[-1]]:
			ans = max(ans, arr[n - i - 1] ^ arr[sBackward[-1]])
			sBackward.pop()

		sBackward.append(n - i - 1)

	return ans

# Driver Code
if __name__ == "__main__":

	arr = [8, 1, 2]
	n = len(arr)
	print(maximumXor(arr, n))

# This code is contributed by
# sanjeev2552
