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.util.Arrays;
import java.util.Deque;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class Main1180C
{
    static Deque<Integer> deq = new LinkedList<>();
    static int n, q, maxv, maxi;
    static Query[] qu;
    static int[] arr;
    public static void main(String[] args) throws IOException
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        StringTokenizer st = new StringTokenizer(br.readLine());
        n = Integer.parseInt(st.nextToken());
        q = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(br.readLine());
        for(int i=0;i<n;++i)
        {
            int v = Integer.parseInt(st.nextToken());
            if(v>maxv)
            {
                maxv = v;
                maxi = i;
            }
            deq.offerLast(v);
        }
        qu = new Query[q];
        for(int i=0;i<q;++i)
        {
            long v = Long.parseLong(br.readLine())-1;
            qu[i] = new Query(v,i);
        }
        Arrays.sort(qu,(o1,o2)->Long.compare(o1.val,o2.val));
        int cur = 0;
        for(int i=0;i<maxi;++i)
        {
            int a = deq.poll();
            int b = deq.poll();
            while(cur<q && i==qu[cur].val)
            {
                qu[cur].a = a;
                qu[cur].b = b;
                ++cur;
            }
            deq.offerFirst(a>b?a:b);
            deq.offerLast(a>b?b:a);
        }
        arr = new int[n-1];
        int a = deq.poll();
        for(int i=0;i<n-1;++i)
            arr[i] = deq.poll();
        for(int i=cur; i<q; ++i)
        {
            long val = qu[i].val - maxi;
            qu[i].a = a;
            qu[i].b = arr[(int)(val%(n-1))];
        }
        Arrays.sort(qu,(o1,o2)->Integer.compare(o1.idx,o2.idx));
        for(Query query : qu)
        {
            bw.write(query.a+" "+query.b+"\n");
        }
        bw.close();
    }

}
class Query
{
    long val;
    int idx;
    int a,b;
    Query(long val, int idx)
    {
        this.val = val; this.idx = idx;
    }
}
