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. 
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 5;
struct Node {
  int value;
  struct Node *next;
} *head = NULL, *tail;
struct Query {
  int index, ans1, ans2;
  long long round;
} queries[maxn];
bool cmp_by_round(const Query &lhs, const Query &rhs) {
  return lhs.round < rhs.round;
}
bool cmp_by_index(const Query &lhs, const Query &rhs) {
  return lhs.index < rhs.index;
}
int a[maxn], b[maxn];
int main() {
  int n, q, maxv = 0;
  scanf("%d%d", &n, &q);
  struct Node *last = NULL;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    struct Node *node = (Node *)malloc(sizeof(Node));
    node->value = a[i];
    node->next = NULL;
    if (head == NULL) head = node;
    if (last) last->next = node;
    last = tail = node;
    if (a[i] > a[maxv]) maxv = i;
  }
  for (int i = 0; i < q; i++) {
    scanf("%lld", &queries[i].round);
    queries[i].index = i;
  }
  sort(queries, queries + q, cmp_by_round);
  int index_query = 0;
  for (int i = 0; i < maxv; i++) {
    struct Node *node1 = (Node *)malloc(sizeof(Node));
    struct Node *node2 = (Node *)malloc(sizeof(Node));
    int v1 = head->value, v2 = head->next->value;
    while (1LL * i == queries[index_query].round - 1) {
      queries[index_query].ans1 = head->value;
      queries[index_query].ans2 = head->next->value;
      index_query++;
    }
    if (v1 < v2) swap(v1, v2);
    node1->value = v1;
    if (head->next->next)
      node1->next = head->next->next;
    else
      node1->next = node2;
    node2->value = v2;
    node2->next = NULL;
    tail->next = node2;
    tail = node2;
    free(head->next);
    free(head);
    head = node1;
  }
  struct Node *node = head;
  for (int i = 0; i < n; i++) {
    b[i] = node->value;
    node = node->next;
  }
  while (index_query < q) {
    long long temp_round = queries[index_query].round - maxv;
    queries[index_query].ans1 = b[0];
    queries[index_query].ans2 = b[int((temp_round - 1) % (n - 1) + 1)];
    index_query++;
  }
  sort(queries, queries + q, cmp_by_index);
  for (int i = 0; i < q; i++)
    printf("%d %d\n", queries[i].ans1, queries[i].ans2);
  return 0;
}
