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: .

The lucky number of the sequence of distinct positive integers x

1, 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, ..., s

r 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}.

題意：給一個長度為 n 的序列，元素不重複，求任意區間中的最大元素與次大元素的異或值，然後再取這些值的最大值，注意區間長度任意。

題解：如果暴力的話是 O(n ^ 2) 的複雜度，顯然不行，所以就需要用單調棧維護一下，O(n)解決，找到第一個比它大的數，看程式碼吧！！

#include <iostream>
#include <cstdio>
#include <stack>
using namespace std;
const int MAX = 1e5+100;
typedef long long ll;
ll a[MAX];
stack<ll> s;
int main(){
	int n;
	scanf("%d",&n);
	for (int i = 0; i < n;i++){
		scanf("%lld",&a[i]);
	}
	ll ans=0;
	for (ll i = n-1; i >= 0;i--){
		while(!s.empty()&&a[s.top()]<a[i]) s.pop();//單調棧找到第一個比它大的數 反向跑
		if(!s.empty()) ans=max(ans,a[s.top()]^a[i]);
		s.push(i);
	}
	for (ll i = 0; i < n;i++){
		while(!s.empty()&&a[s.top()]<a[i]) s.pop();//單調棧找到第一個比它大的數 正向跑
		if(!s.empty()) ans=max(ans,a[s.top()]^a[i]);
		s.push(i);
	}
// 必須兩個方向跑，要不然會有遺漏
	printf("%lld\n",ans);
	return 0;
}