#include <bits/stdc++.h>
using namespace std;
long long const INF = INT_MAX;
const long long MOD = 998244353;
const int N = 100020;
void SpeedForce() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int simple(int n) {
  int i, sq;
  if (n % 2 == 0) {
    return 0;
  }
  sq = (int)sqrt(n);
  for (i = 3; i <= sq; i++) {
    if ((n % i) == 0) {
      return 0;
    }
  }
  return 1;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long is_prime(long long n) {
  if (n <= 1 || n > 3 && (n % 2 == 0 || n % 3 == 0)) return 0;
  for (long long i = 5, t = 2; i * i <= n; i += t, t = 6 - t)
    if (n % i == 0) return 0;
  return 1;
}
long long binpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) {
      res = (res * x) % MOD;
    }
    x = (x * x) % MOD;
    n /= 2;
  }
  return res;
}
int a[N];
pair<int, int> o[N];
int q[N << 1];
void solve() {
  int n, m, k;
  int i, j, p;
  int l, r;
  long long x;
  cin >> n >> m;
  for (i = k = l = r = 0; i < n; i++) {
    cin >> a[i];
    k = max(k, a[i]);
    q[r++] = a[i];
  }
  for (i = 0; i < n; i++)
    if (a[i] == k) {
      p = i;
      break;
    }
  for (i = 0; i < p; i++) {
    o[i] = make_pair(q[l], q[l + 1]);
    q[r++] = min(q[l], q[l + 1]);
    q[l + 1] = max(q[l], q[l + 1]);
    l++;
  }
  while (m--) {
    cin >> x;
    x--;
    if (x < p)
      cout << o[x].first << " " << o[x].second << '\n';
    else
      cout << a[p] << " " << q[(x - p) % (n - 1) + l + 1] << '\n';
  }
}
int main() {
  SpeedForce();
  int T = 1;
  while (T--) {
    solve();
  }
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long const INF = INT_MAX;
const long long MOD = 998244353;
const int N = 100020;
void SpeedForce() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int simple(int n) {
  int i, sq;
  if (n % 2 == 0) {
    return 0;
  }
  sq = (int)sqrt(n);
  for (i = 3; i <= sq; i++) {
    if ((n % i) == 0) {
      return 0;
    }
  }
  return 1;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long is_prime(long long n) {
  if (n <= 1 || n > 3 && (n % 2 == 0 || n % 3 == 0)) return 0;
  for (long long i = 5, t = 2; i * i <= n; i += t, t = 6 - t)
    if (n % i == 0) return 0;
  return 1;
}
long long binpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) {
      res = (res * x) % MOD;
    }
    x = (x * x) % MOD;
    n /= 2;
  }
  return res;
}
int a[N];
pair<int, int> o[N];
int q[N << 1];
void solve() {
  int n, m, k;
  int i, j, p;
  int l, r;
  long long x;
  cin >> n >> m;
  for (i = k = l = r = 0; i < n; i++) {
    cin >> a[i];
    k = max(k, a[i]);
    q[r++] = a[i];
  }
  for (i = 0; i < n; i++)
    if (a[i] == k) {
      p = i;
      break;
    }
  for (i = 0; i < p; i++) {
    o[i] = make_pair(q[l], q[l + 1]);
    q[r++] = min(q[l], q[l + 1]);
    q[l + 1] = max(q[l], q[l + 1]);
    l++;
  }
  while (m--) {
    cin >> x;
    x--;
    if (x < p)
      cout << o[x].first << " " << o[x].second << '\n';
    else
      cout << a[p] << " " << q[(x - p) % (n - 1) + l + 1] << '\n';
  }
}
int main() {
  SpeedForce();
  int T = 1;
  while (T--) {
    solve();
  }
}

```