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

 f(b) = \begin{cases} b[1] &   if  m = 1 \\\ f(b[1] ⊕ b[2],b[2] ⊕ b[3],...,b[m-1] ⊕ b[m]) &   otherwise, \end{cases}  

where ⊕ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

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 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}, …, a_r.

Input

The first line contains a single integer n (1 ≤ n ≤ 5000) — the length of a.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30}-1) — the elements of the array.

The third line contains a single integer q (1 ≤ q ≤ 100 000) — the number of queries.

Each of the next q lines contains a query represented as two integers l, r (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 [3,6], for second query — [2,5], for third — [3,4], for fourth — [1,2].
import java.util.*;
import java.io.*;
//import java.text.*;

public class Main{
    final long MOD = (long)1e9+7, IINF = (long)1e19;
    final int MAX = (int)1e6+10, MX = (int)1e7+1, INF = (int)1e9;
//    DecimalFormat df = new DecimalFormat("0.00000000");
//    final double EPS = 1e-8;
    FastReader in;
    PrintWriter out;
    boolean multipleTC = false;
    
    public static void main(String[] args) throws Exception{new Main().run();};
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        for(int i = 1, t = (multipleTC)?ni():1; i<=t; i++)solve(i);
        out.flush();    
        out.close();
    }
    
    void solve(int TC){
        int n = ni();
        int[] a = ia(0,n);
        int[][] dp = new int[n][n], ans = new int[n][n];
        for(int i = 0; i< n; i++){
            dp[0][i] = a[i];
            ans[0][i] = a[i];
        }
        for(int i = 1; i< n; i++){
            for(int j = 0; i+j<n; j++){
                dp[i][j] = dp[i-1][j]^dp[i-1][j+1];
                ans[i][j] = Math.max(dp[i][j], Math.max(ans[i-1][j], ans[i-1][j+1]));
            }
        }
        for(int q = ni(); q>0; q--){
            int l = ni()-1, r = ni()-1;
            int i = r-l;
            pn(ans[i][l]);
        }
    }
    
    long mul(long a, long b){
        if(a>=MOD)a%=MOD;
        if(b>=MOD)b%=MOD;
        a*=b;
        return (a>=MOD)?a%MOD:a;
    }
    
    int[] reverse(int[] a){
        int[] o = new int[a.length];
        for(int i = 0; i< a.length; i++)o[i] = a[a.length-i-1];
        return o;   
    }
    
    int[] sort(int[] a){
        if(a.length==1)return a;
        int mid = a.length/2;
        int[] b = sort(Arrays.copyOfRange(a,0,mid)), c = sort(Arrays.copyOfRange(a,mid,a.length));
        for(int i = 0, j = 0, k = 0; i< a.length; i++){
            if(j<b.length && k<c.length){
                if(b[j]<c[k])a[i] = b[j++];
                else a[i] = c[k++];
            }else if(j<b.length)a[i] = b[j++];
            else a[i] = c[k++];
        }
        return a;
    }
    
    long[] sort(long[] a){
        if(a.length==1)return a;
        int mid = a.length/2;
        long[] b = sort(Arrays.copyOfRange(a,0,mid)), c = sort(Arrays.copyOfRange(a,mid,a.length));
        for(int i = 0, j = 0, k = 0; i< a.length; i++){
            if(j<b.length && k<c.length){
                if(b[j]<c[k])a[i] = b[j++];
                else a[i] = c[k++];
            }else if(j<b.length)a[i] = b[j++];
            else a[i] = c[k++];
        }
        return a;
    }
    

    int[] ia(int ind,int n){
        int[] out = new int[ind+n];
        for(int i = 0; i< n; i++)out[ind+i] = ni();
        return out;
    }

    long[] la(int ind, int n){
        long[] out = new long[ind+n];
        for(int i = 0; i< n; i++)out[ind+i] = nl();
        return out;
    }

    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }
            return str;
        }
    }
}