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CF 280B -——Maximum Xor Secondary(单调栈)
阅读量:517 次
发布时间:2019-03-07

本文共 1700 字,大约阅读时间需要 5 分钟。

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 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

55 2 1 4 3

Output

7

Input

59 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}.

题意:给一个长度为 n 的序列,元素不重复,求任意区间中的最大元素与次大元素的异或值,然后再取这些值的最大值,注意区间长度任意。

题解:如果暴力的话是 O(n ^ 2) 的复杂度,显然不行,所以就需要用单调栈维护一下,O(n)解决,找到第一个比它大的数,看代码吧!!

#include 
#include
#include
using namespace std;const int MAX = 1e5+100;typedef long long ll;ll a[MAX];stack
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()]

 

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