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 bb of length mm we define the function ff as

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

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

这题就是在异或操作下的区间最大值。

n《=5000 可以用区间DP做。

dp[i][j] 表示长度为i 起始点是j 的区间最大值。

 #include<bits/stdc++.h>
 using namespace std;

 typedef long long LL ;
 const int maxn = 5e3 + ;
 int dp[maxn][maxn];

 int main() {
     int n;
     scanf("%d", &n );
     for (int i =  ; i <= n ; i++)
         scanf("%d", &dp[][i]);
     for (int i =  ; i <= n ; i++ )
         for (int j =  ; j <= n - i +  ; j++)
             dp[i][j] = dp[i - ][j] ^ dp[i - ][j + ];
     for (int i =  ; i <= n ; i++)
         for (int j =  ; j <= n - i +  ; j++)
             dp[i][j] = max(dp[i][j], max(dp[i - ][j], dp[i - ][j + ]));
     int q;
     scanf("%d", &q);
     while(q--) {
         int x, y;
         scanf("%d%d", &x, &y);
         int len = y - x + ;
         printf("%d\n", dp[len][x]);
     }
     return ;
 }