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

D. 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.

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 — $[1,2]$.

给n个数，询问q次，每次询问给出l,r.   [l,r]区间求异或最大值为多少。

与其他语言不同，c语言和c++语言的异或不用X0R；而用‘^’,这在其他语言表示乘方。

此题需要记忆化两次。区间动态规划。

#include
#define MAX 5005 
#define  ll long long
using namespace std;
ll N,Q;
ll a[MAX],dp[MAX][MAX],ans[MAX][MAX];
int main()
{
std::ios::sync_with_stdio(0); 
cin.tie(0);
cout.tie(0); 
cin>>N;
for(ll i=1;i<=N;i++)
    {cin>>a[i]; dp[i][i]=a[i]; ans[i][i]=a[i];//类似杨辉三角初始化 
}
for(ll t=1;t    for(ll i=1;i+t<=N;i++)//枚举开始位置 i+t为结束位置
   {
      ll j=i+t;
      dp[i][j]=dp[i+1][j]^dp[i][j-1];
  ans[i][j]=max(dp[i][j],max(ans[i+1][j],ans[i][j-1]));//自己画个矩阵就明白了 
} 
cin>>Q;
while(Q--)
{
ll l,r;
cin>>l>>r;
cout< } 
return 0;
}