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

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 $a^{}$

The first line contains a single integer $n$

Each of the next $q$lines contains a query represented as two integers $l$

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

3
8 4 1
2
2 3
1 2

5
12

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

60
30
12
3

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 $[$

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 5010

int n,q,l,r;
__int64 ans[N][N],dp[N][N],a[N];

int main()
{
    while(~scanf("%d",&n))
    {
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=n; i++)
        {
            scanf("%I64d",&a[i]);
            ans[i][i]=a[i];
            dp[i][i]=a[i];
        }
        for(int l=1; l<n; l++)
        {
            for(int i=1; i+l<=n; i++)
            {
                int j=l+i;
                ans[i][j]=ans[i][j-1]^ans[i+1][j];
                dp[i][j]=max(ans[i][j],max(dp[i+1][j],dp[i][j-1]));
            }
        }
        scanf("%d",&q);
        while(q--)
        {
            scanf("%d%d",&l,&r);
            printf("%I64d\n",dp[l][r]);
        }
    }
    return 0;
}