Codeforces Round #483 (Div. 2) D 、XOR-pyramid（DP）984D

D. XOR-pyramid

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

For an array $b$ of length $m$ we define the function $f$ as

$f\left(b\right)=\{\begin{array}{cc}b\left[1\right]& \text{if}m=1\\ f\left(b\right[1]\oplus b[2],b[2]\oplus b[3],\dots ,b[m-1]\oplus b[m\left]\right)& \text{otherwise,}\end{array}$

where $\oplus$ is bitwise exclusive OR.

For example, $f(1,2,4,8)=f(1\oplus 2,2\oplus 4,4\oplus 8)=f(3,6,12)=f(3\oplus 6,6\oplus 12)=f(5,10)=f(5\oplus 10)=f\left(15\right)=15$

You are given an array $a$ and a few queries. Each query is represented as two integers $l$ and $r$. The answer is the maximum value of $f$ on all continuous subsegments of the array $a_l,a_{l+1},\dots ,a_r$.

Input

The first line contains a single integer $n$ ($1\le n\le 5000$) — the length of $a$.

The second line contains $n$ integers $a_1,a_2,\dots ,a_n$ ($0\le a_i\le 2^{30}-1$) — the elements of the array.

The third line contains a single integer $q$ ($1\le q\le 100000$) — the number of queries.

Each of the next $q$ lines contains a query represented as two integers $l$, $r$ ($1\le l\le r\le n$).

Output

Print $q$ lines — the answers for the queries.

Examples

input

Copy

3
8 4 1
2
2 3
1 2

output

Copy

5
12

input

Copy

6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2

output

Copy

60
30
12
3

Note

In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.

In second sample, optimal segment for first query are $[3,6]$, for second query — $[2,5]$, for third — $[3,4]$, for fourth — $[1,2]$.

一、原题地址

    传送门

二、大致题意

    给出n个数，查询在（l,r）段上的最大的异或值。

三、思路

    将整张图更新完成，然后O（n）查询。

    状态的转移方程为m[i][j]=m[i][j-1]^m[i+1][j].

四、代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<stack>
using namespace std;
const int inf=0x3f3f3f3f;
long long  gcd(long long  a,long long  b){ return a == 0 ? b : gcd(b % a, a); } 


int n;
int m[5005][5005],ans[5005][5005];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&m[i][i]);
		ans[i][i]=m[i][i];
	}
	for(int len=0;len<=n-2;len++)
	{
		for(int l=1;l+len+1<=n;l++)
		{
			int r=l+len+1;
			m[l][r]=m[l][r-1]^m[l+1][r];
			ans[l][r]=max(m[l][r],max(ans[l][r-1],ans[l+1][r]));
		}
	}
	int q;
	scanf("%d",&q);
	while(q--)
	{
		int l,r;
		scanf("%d%d",&l,&r);
		printf("%d\n",ans[l][r]);
	}
	getchar();
	getchar();
}