XOR-pyramid(Codeforces Round #483 (Div. 2) )(dp)

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 — $[1,2]$

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这是一道dp题，使用两个数组，一个数组为f数组，一个数组为ans数组，f数组枚举出所有情况，ans通过动态转移方程来求出结果。

动态转移方程为：ans[l][r]=max(f[l][r],max(ans[l][r-1],ans[l+1][r]));

本题代码如下：

#include<bits/stdc++.h>

using namespace std;

const int maxn=5e3+10;

typedef long long LL;

LL ans[maxn][maxn],f[maxn][maxn];

int main()
{
	LL n;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>f[i][i];
		ans[i][i]=f[i][i];
	}	
	cout<<endl;
	for(int len=2;len<=n;len++){
		for(int l=1,r;(r=l+len-1)<=n;l++){
			f[l][r]=f[l][r-1]^f[l+1][r];
			ans[l][r]=max(f[l][r],max(ans[l][r-1],ans[l+1][r]));
		}
	}
	// for(int i=1;i<=n;i++){
	// 	for(int j=1;j<=n;j++){
	// 		printf("%5d",f[i][j]);
	// 	}
	// 	cout<<endl;
	// }
	// cout<<endl;
	// for(int i=1;i<=n;i++){
	// 	for(int j=1;j<=n;j++){
	// 		printf("%5d",ans[i][j]);
	// 	}
	// 	cout<<endl;
	// }
	LL q,l,r;
	cin>>q;
	while(q--){
		cin>>l>>r;
		cout<<ans[l][r]<<endl;
	}
	return 0;
}