B. XOR-pyramid+区间dp

B. 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.

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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 — $[1,2]$.

#define happy

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define all(a) (a).begin(),(a).end()
#define pll pair<ll,ll>
#define vi vector<int>
#define pb push_back
const int inf=0x3f3f3f3f;
ll rd(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int N=5e3+10;
int n,m,f[N][N];

int main(){
#ifdef happy
    freopen("in.txt","r",stdin);
#endif
    int n=rd();
    rep(i,1,n)
    f[i][i]=rd();
    int i,j;
    rep(k,1,n-1)
    for(i=1;(j=i+k)<=n;i++)
    f[i][j]=f[i][j-1]^f[i+1][j];
    rep(k,1,n-1)
    for(i=1;(j=i+k)<=n;i++)
    f[i][j]=max(f[i][j],max(f[i][j-1],f[i+1][j]));
    int m=rd();
    while(m--){
        i=rd(),j=rd();
        printf("%d\n",f[i][j]);
    }
}