CodeForces - 983B XOR-pyramid

 

   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 f as

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

where 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)=15

You are given an array al,al+1,…,ar.

Input

The first line contains a single integer a.

The second line contains 0≤ai≤230−1) — the elements of the array.

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

Each of the next 1≤l≤r≤n).

Output

Print q 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 [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;
}