Recently, on the course of algorithms and data structures, Valeriy learned how to use a deque. He built a deque filled with n elements. The i-th element is a_i (i = 1, 2, …, n). He gradually takes the first two leftmost elements from the deque (let's call them A and B, respectively), and then does the following: if A > B, he writes A to the beginning and writes B to the end of the deque, otherwise, he writes to the beginning B, and A writes to the end of the deque. We call this sequence of actions an operation.

For example, if deque was [2, 3, 4, 5, 1], on the operation he will write B=3 to the beginning and A=2 to the end, so he will get [3, 4, 5, 1, 2].

The teacher of the course, seeing Valeriy, who was passionate about his work, approached him and gave him q queries. Each query consists of the singular number m_j (j = 1, 2, …, q). It is required for each query to answer which two elements he will pull out on the m_j-th operation.

Note that the queries are independent and for each query the numbers A and B should be printed in the order in which they will be pulled out of the deque.

Deque is a data structure representing a list of elements where insertion of new elements or deletion of existing elements can be made from both sides.

Input

The first line contains two integers n and q (2 ≤ n ≤ 10^5, 0 ≤ q ≤ 3 ⋅ 10^5) — the number of elements in the deque and the number of queries. The second line contains n integers a_1, a_2, ..., a_n, where a_i (0 ≤ a_i ≤ 10^9) — the deque element in i-th position. The next q lines contain one number each, meaning m_j (1 ≤ m_j ≤ 10^{18}).

Output

For each teacher's query, output two numbers A and B — the numbers that Valeriy pulls out of the deque for the m_j-th operation.

Examples

Input


5 3
1 2 3 4 5
1
2
10


Output


1 2
2 3
5 2


Input


2 0
0 0


Output

Note

Consider all 10 steps for the first test in detail:
  1. [1, 2, 3, 4, 5] — on the first operation, A and B are 1 and 2, respectively.

So, 2 we write to the beginning of the deque, and 1 — to the end.

We get the following status of the deque: [2, 3, 4, 5, 1].

  2. [2, 3, 4, 5, 1] ⇒ A = 2, B = 3.
  3. [3, 4, 5, 1, 2]
  4. [4, 5, 1, 2, 3]
  5. [5, 1, 2, 3, 4]
  6. [5, 2, 3, 4, 1]
  7. [5, 3, 4, 1, 2]
  8. [5, 4, 1, 2, 3]
  9. [5, 1, 2, 3, 4]
  10. [5, 2, 3, 4, 1] ⇒ A = 5, B = 2. 
import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.util.*;
import java.util.concurrent.LinkedBlockingDeque;


public class scratch_25 {
    // int count=0;
    //static long count=0;
    static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        /**
         * call this method to initialize reader for InputStream
         */
        static void init(InputStream input) {
            reader = new BufferedReader(
                    new InputStreamReader(input));
            tokenizer = new StringTokenizer("");
        }

        /**
         * get next word
         */
        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                //TODO add check for eof if necessary
                tokenizer = new StringTokenizer(
                        reader.readLine());
            }
            return tokenizer.nextToken();
        }

        static int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        static double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        static long nextLong() throws IOException {
            return Long.parseLong(next());
        }
    }





    static class Pair implements Comparable<Pair>{
        long x;
        long y;
        public Pair(long x,long y){
            this.x=x;
            this.y=y;
        }
        @Override
        public int compareTo(Pair o){
            return (int)this.y-(int)o.y;


        }
        @Override
        public boolean equals(Object me) {
            Pair binMe = (Pair)me;
            if(this.x==binMe.x && this.y==binMe.y)
                return true;
            else
                return false;
        }

        @Override
        public int hashCode() {
            return (int)this.x + (int)this.y;
        }

        @Override
        public String toString() {
            return x+" "+y;
        }


    }






    public static void main(String[] args)  throws IOException {
        Reader.init(System.in);
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
        int n= Reader.nextInt();
        int q= Reader.nextInt();
        Deque<Integer>ar= new LinkedList<>();
        int max=0;
        int ind=0;
        for (int i = 0; i <n ; i++) {
         //   System.out.println("lol");
            int x= Reader.nextInt();
            if(x>max){
                ind=i;
                max=x;
            }
            ar.add(x);
        }
        int que[][]= new int[ind][2];
        for (int i = 0; i <ind ; i++) {
         //   System.out.println("fuck");
            int a1=ar.remove();
            int a2=ar.remove();
            int ma= Math.max(a1,a2);
            int mi= Math.min(a1,a2);
            ar.addFirst(ma);
            ar.addLast(mi);
            que[i][0]=a1;
            que[i][1]=a2;
        }
     //   System.out.println(ar);
        int v= ar.size();
        ArrayList<Integer>x= new ArrayList<>();
        for (int i = 0; i <v ; i++) {
            x.add(ar.remove());
        }

        for (int i = 0; i <q ; i++) {
            long w= Reader.nextLong();
            if(w<=ind){
               out.append(que[(int)w-1][0]+" "+que[(int)w-1][1]+"\n");
            }
            else{
                long c= w-(ind+1);
                long g= c%(n-1);
                int a=(int)g;
                out.append(max+" "+x.get(a+1)+"\n");
            }

        }












        out.flush();
        out.close();



    }



    public static void bfs(int x,int y, int n,char [][]ar,int done[][]){
        LinkedList<int[]>stk= new LinkedList<>();
        int a[]={x,y};
        stk.addFirst(a);
        while(!stk.isEmpty()){
            int b[]=stk.removeFirst();
            if(done[b[0]][b[1]]==1){
                continue;
            }
            done[b[0]][b[1]]=1;
            int arr[][]={{1,0},{-1,0},{0,1},{0,-1}};
            for (int i = 0; i <arr.length ; i++) {
                if(!(b[0]+arr[i][0]>=n || b[1]+arr[i][1]>=n || b[0]+arr[i][0]<0 || b[1]+arr[i][1]<0 ||  ar[b[0]+arr[i][0]][b[1]+arr[i][1]]=='1')){
                    int r[]= {b[0]+arr[i][0], b[1]+arr[i][1]};
                    stk.addFirst(r);
                    // continue;
                }
            }
        }
    }




























    public static ArrayList<Integer> Sieve(int n) {
        boolean arr[]= new boolean [n+1];
        Arrays.fill(arr,true);
        arr[0]=false;
        arr[1]=false;
        for (int i = 2; i*i <=n ; i++) {
            if(arr[i]){
                for (int j = 2; j <=n/i ; j++) {
                    int u= i*j;
                    arr[u]=false;
                }}
        }
        ArrayList<Integer> ans= new ArrayList<>();
        for (int i = 0; i <n+1 ; i++) {
            if(arr[i]){
                ans.add(i);
            }
        }
        return ans;
    }




    static long power( long x, long y, long p)
    {

        long res = 1;


        x = x % p;

        if (x == 0) return 0;

        while (y > 0)
        {

            if((y & 1)==1)
                res = (res * x) % p;


            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }

    public static long ceil_div(long a, long b){
        return (a+b-1)/b;
    }

    static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    static long lcm(long a, long b)
    {
        return (a*b)/gcd(a, b);
    }

}