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 inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long _INF = 0xc0c0c0c0c0c0c0c0;
const long long mod = (int)1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long ksm(long long a, long long b, long long mod) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long inv2(long long a, long long mod) { return ksm(a, mod - 2, mod); }
void exgcd(long long a, long long b, long long &x, long long &y, long long &d) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, y, x, d);
    y -= x * (a / b);
  }
}
const int MAX_N = 100025;
int STA[MAX_N], END[MAX_N], STA_[MAX_N], END_[MAX_N], arr[MAX_N], ANS[MAX_N][3],
    ANS_[MAX_N][3];
queue<int> q, q_;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
    if (i != 1 && i != n) q.push(arr[i]);
  }
  STA[0] = arr[1];
  END[0] = arr[n];
  for (int i = 1; i <= n; ++i) {
    int tmp = STA[i - 1];
    int tmp_ = q.front();
    q.pop();
    q.push(END[i - 1]);
    ANS[i][0] = tmp;
    ANS[i][1] = tmp_;
    if (tmp > tmp_) {
      STA[i] = tmp;
      END[i] = tmp_;
    } else {
      STA[i] = tmp_;
      END[i] = tmp;
    }
  }
  STA_[0] = STA[n];
  END_[0] = END[n];
  while (!q.empty()) {
    int tmp = q.front();
    q.pop();
    q_.push(tmp);
  }
  for (int i = 1; i < n; ++i) {
    int tmp = STA_[i - 1];
    int tmp_ = q_.front();
    q_.pop();
    q_.push(END_[i - 1]);
    ANS_[i][0] = tmp;
    ANS_[i][1] = tmp_;
    if (tmp > tmp_) {
      STA_[i] = tmp;
      END_[i] = tmp_;
    } else {
      STA_[i] = tmp_;
      END_[i] = tmp;
    }
  }
  if (n == 2) {
    for (int i = 1; i <= m; ++i) {
      long long ask;
      scanf("%lld", &ask);
      if (ask == 1) {
        printf("%d %d\n", arr[1], arr[2]);
      } else {
        printf("%d %d\n", max(arr[1], arr[2]), min(arr[1], arr[2]));
      }
    }
  } else {
    for (int i = 1; i <= m; ++i) {
      long long ask;
      scanf("%lld", &ask);
      if (ask <= n) {
        printf("%d %d\n", ANS[ask][0], ANS[ask][1]);
      } else {
        ask -= n;
        ask %= (n - 1);
        if (ask == 0) ask = n - 1;
        printf("%d %d\n", ANS_[ask][0], ANS_[ask][1]);
      }
    }
  }
  return 0;
}
