CodeForces - 983B XOR-pyramid

Posted 2020-11-06 蒟蒻JHY

 

   设f[l,r]为区间[l,r]的f值，那么可以很容易的发现f[l,r] = f[l+1,r] ^ f[l,r-1] (l<r)  or  a[l] (l==r)，这个在纸上画一画就发现了。

 

   于是就成了一个SB题了2333

 

Discription

For an array bb of length mm we define the function ff as

f(b)={b[1]f(b[1]⊕b[2],b[2]⊕b[3],…,b[m?1]⊕b[m])if m=1otherwise,f(b)={b[1]if m=1f(b[1]⊕b[2],b[2]⊕b[3],…,b[m?1]⊕b[m])otherwise,

where ⊕⊕ is bitwise exclusive OR.

For example, f(1,2,4,8)=f(1⊕2,2⊕4,4⊕8)=f(3,6,12)=f(3⊕6,6⊕12)=f(5,10)=f(5⊕10)=f(15)=15f(1,2,4,8)=f(1⊕2,2⊕4,4⊕8)=f(3,6,12)=f(3⊕6,6⊕12)=f(5,10)=f(5⊕10)=f(15)=15

You are given an array aa and a few queries. Each query is represented as two integers ll and rr. The answer is the maximum value of ff on all continuous subsegments of the array al,al+1,…,aral,al+1,…,ar.

Input

The first line contains a single integer nn (1≤n≤50001≤n≤5000) — the length of aa.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤230?10≤ai≤230?1) — the elements of the array.

The third line contains a single integer qq (1≤q≤1000001≤q≤100000) — the number of queries.

Each of the next qq lines contains a query represented as two integers ll, rr (1≤l≤r≤n1≤l≤r≤n).

Output

Print qq lines — the answers for the queries.

Examples

Input

3
8 4 1
2
2 3
1 2

Output

5
12

Input

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

Output

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][3,6], for second query — [2,5][2,5], for third — [3,4][3,4], for fourth — [1,2][1,2].

 

 

 

 

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=5005;
int f[maxn][maxn],ans[maxn][maxn],n,Q,L,R;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&f[i][i]);
	for(int l=1;l<n;l++)
	    for(int i=1,j=l+1;j<=n;i++,j++){
	    	f[i][j]=f[i+1][j]^f[i][j-1];
	    	ans[i][j]=max(f[i][j],max(ans[i+1][j],ans[i][j-1]));
		}
	
	scanf("%d",&Q);
	while(Q--) scanf("%d%d",&L,&R),printf("%d\n",ans[L][R]);
	
	return 0;
}