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;
struct myp {
  int f;
  int s;
};
struct edges {
  int v;
  bool check1;
};
int dx[5] = {-1, 0, 0, 1, 0};
myp saveme[100002];
int dy[5] = {0, 1, -1, 0, 0};
int len[1008][1008];
bool checktui[1008][1008];
int accum[1008][1008];
bool alpha[28];
int countval[200008];
int chooseval[200008];
long long first1[10009];
long long first2[10009];
long long save[500009];
long long second[1000009];
int deg[10000];
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
bool isok(int n, int m, int x, int y, char** a) {
  if (x >= 0 && x < n && y >= 0 && y < m) {
    return true;
    return false;
  }
  return false;
}
int gcd(int num, int denum) {
  if (num == 0)
    return denum;
  else
    return gcd(denum % num, num);
}
int get(long long x, int& finalzero) {
  int len = 0;
  while (x != 0) {
    len += 1;
    if (x % 2 == 0) {
      if (len <= 30) finalzero = len;
    }
    x /= 2;
  }
  return len;
}
int myabs(int a) {
  if (a >= 0)
    return a;
  else
    return -a;
}
void resetval() {
  for (int i = 0; i < 200008; i++) {
    countval[i] = 0;
    chooseval[i] = 0;
  }
}
int main() {
  int q;
  long long n;
  cin >> n;
  cin >> q;
  deque<int> a;
  int x;
  for (int i = 0; i < n; i++) {
    cin >> x;
    a.push_back(x);
  }
  saveme[1].f = a[0];
  saveme[1].s = a[1];
  for (int i = 2; i < n + 1; i++) {
    if (a[0] > a[1]) {
      int tmp = a[0];
      int tmp1 = a[1];
      a.pop_front();
      a.pop_front();
      a.push_front(tmp);
      a.push_back(tmp1);
    } else {
      int tmp = a[0];
      int tmp1 = a[1];
      a.pop_front();
      a.pop_front();
      a.push_front(tmp1);
      a.push_back(tmp);
    }
    saveme[i].f = a[0];
    saveme[i].s = a[1];
  }
  for (int i = 0; i < q; i++) {
    long long m;
    cin >> m;
    if (m >= n)
      cout << saveme[n].f << " " << a[(m - n) % (n - 1) + 1] << endl;
    else
      cout << saveme[m].f << " " << saveme[m].s << endl;
  }
}
