#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
std::pair<int, int> DR[] = {{-1, 0}, {0, 1},   {1, 0}, {0, -1},
                            {-1, 1}, {-1, -1}, {1, 1}, {1, -1}};
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  ~debug() { cerr << endl; }
  template <class c>
  typename enable_if<sizeof dud<c>(0) != 1, debug&>::type operator<<(c i) {
    cerr << boolalpha << i;
    return *this;
  }
  template <class c>
  typename enable_if<sizeof dud<c>(0) == 1, debug&>::type operator<<(c i) {
    return *this << range(begin(i), end(i));
  }
  template <class c, class b>
  debug& operator<<(pair<b, c> d) {
    return *this << "(" << d.first << ", " << d.second << ")";
  }
  template <class c>
  debug& operator<<(rge<c> d) {
    *this << "[";
    for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
    return *this << "]";
  }
};
long long gcd(long long a, long long b) {
  if (b) return gcd(b, a % b);
  return a;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
deque<long long> dq;
long long n, q, x, mx;
pair<long long, long long> ans[1 << 19];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> x;
    mx = max(mx, x);
    dq.push_back(x);
  }
  long long cnt = 0;
  while (1) {
    if (dq.front() == mx) break;
    cnt++;
    long long x = dq.front();
    dq.pop_front();
    long long y = dq.front();
    dq.pop_front();
    ans[cnt] = make_pair(x, y);
    if (y > x) swap(x, y);
    dq.push_back(y);
    dq.push_front(x);
  }
  vector<long long> vec;
  dq.pop_front();
  for (auto it : dq) {
    vec.push_back(it);
  }
  while (q--) {
    long long x;
    cin >> x;
    if (x <= cnt) {
      cout << ans[x].first << ' ' << ans[x].second << '\n';
    } else {
      x -= cnt;
      x--;
      cout << mx << ' ' << vec[x % (long long)((vec).size())] << '\n';
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
std::pair<int, int> DR[] = {{-1, 0}, {0, 1},   {1, 0}, {0, -1},
                            {-1, 1}, {-1, -1}, {1, 1}, {1, -1}};
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  ~debug() { cerr << endl; }
  template <class c>
  typename enable_if<sizeof dud<c>(0) != 1, debug&>::type operator<<(c i) {
    cerr << boolalpha << i;
    return *this;
  }
  template <class c>
  typename enable_if<sizeof dud<c>(0) == 1, debug&>::type operator<<(c i) {
    return *this << range(begin(i), end(i));
  }
  template <class c, class b>
  debug& operator<<(pair<b, c> d) {
    return *this << "(" << d.first << ", " << d.second << ")";
  }
  template <class c>
  debug& operator<<(rge<c> d) {
    *this << "[";
    for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
    return *this << "]";
  }
};
long long gcd(long long a, long long b) {
  if (b) return gcd(b, a % b);
  return a;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
deque<long long> dq;
long long n, q, x, mx;
pair<long long, long long> ans[1 << 19];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> x;
    mx = max(mx, x);
    dq.push_back(x);
  }
  long long cnt = 0;
  while (1) {
    if (dq.front() == mx) break;
    cnt++;
    long long x = dq.front();
    dq.pop_front();
    long long y = dq.front();
    dq.pop_front();
    ans[cnt] = make_pair(x, y);
    if (y > x) swap(x, y);
    dq.push_back(y);
    dq.push_front(x);
  }
  vector<long long> vec;
  dq.pop_front();
  for (auto it : dq) {
    vec.push_back(it);
  }
  while (q--) {
    long long x;
    cin >> x;
    if (x <= cnt) {
      cout << ans[x].first << ' ' << ans[x].second << '\n';
    } else {
      x -= cnt;
      x--;
      cout << mx << ' ' << vec[x % (long long)((vec).size())] << '\n';
    }
  }
  return 0;
}

```