#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 7340033;
const double eps = 1e-8;
const double PI = acos(-1);
template <class T, class S>
inline void add(T &a, S b) {
  a += b;
  if (a >= mod) a -= mod;
}
template <class T, class S>
inline void sub(T &a, S b) {
  a -= b;
  if (a < 0) a += mod;
}
template <class T, class S>
inline bool chkmax(T &a, S b) {
  return a < b ? a = b, true : false;
}
template <class T, class S>
inline bool chkmin(T &a, S b) {
  return a > b ? a = b, true : false;
}
int n, q, a[N];
int mn, mx, maxVal;
int ans[N][2];
int who[N];
set<pair<int, int> > Set;
int main() {
  scanf("%d%d", &n, &q);
  mn = 1, mx = n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    Set.insert(make_pair(i, a[i]));
    chkmax(maxVal, a[i]);
  }
  int cnt = 0;
  while (Set.begin()->second != maxVal) {
    cnt++;
    auto it = Set.begin();
    pair<int, int> tmp1 = *it;
    ans[cnt][0] = it->second;
    ++it;
    ans[cnt][1] = it->second;
    pair<int, int> tmp2 = *it;
    Set.erase(tmp1);
    Set.erase(tmp2);
    if (tmp1.second > tmp2.second) {
      tmp2.first = ++mx;
      Set.insert(tmp1);
      Set.insert(tmp2);
    } else {
      tmp1.first = ++mx;
      Set.insert(tmp1);
      Set.insert(tmp2);
    }
  }
  auto it = Set.begin();
  ++it;
  int p = 1;
  while (it != Set.end()) {
    who[p] = it->second;
    ++it;
    ++p;
  }
  while (q--) {
    long long m;
    scanf("%lld", &m);
    if (m <= cnt)
      printf("%d %d\n", ans[m][0], ans[m][1]);
    else {
      m -= cnt;
      m %= (n - 1);
      if (!m) m += n - 1;
      printf("%d %d\n", maxVal, who[m]);
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int N = 3e5 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 7340033;
const double eps = 1e-8;
const double PI = acos(-1);
template <class T, class S>
inline void add(T &a, S b) {
  a += b;
  if (a >= mod) a -= mod;
}
template <class T, class S>
inline void sub(T &a, S b) {
  a -= b;
  if (a < 0) a += mod;
}
template <class T, class S>
inline bool chkmax(T &a, S b) {
  return a < b ? a = b, true : false;
}
template <class T, class S>
inline bool chkmin(T &a, S b) {
  return a > b ? a = b, true : false;
}
int n, q, a[N];
int mn, mx, maxVal;
int ans[N][2];
int who[N];
set<pair<int, int> > Set;
int main() {
  scanf("%d%d", &n, &q);
  mn = 1, mx = n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    Set.insert(make_pair(i, a[i]));
    chkmax(maxVal, a[i]);
  }
  int cnt = 0;
  while (Set.begin()->second != maxVal) {
    cnt++;
    auto it = Set.begin();
    pair<int, int> tmp1 = *it;
    ans[cnt][0] = it->second;
    ++it;
    ans[cnt][1] = it->second;
    pair<int, int> tmp2 = *it;
    Set.erase(tmp1);
    Set.erase(tmp2);
    if (tmp1.second > tmp2.second) {
      tmp2.first = ++mx;
      Set.insert(tmp1);
      Set.insert(tmp2);
    } else {
      tmp1.first = ++mx;
      Set.insert(tmp1);
      Set.insert(tmp2);
    }
  }
  auto it = Set.begin();
  ++it;
  int p = 1;
  while (it != Set.end()) {
    who[p] = it->second;
    ++it;
    ++p;
  }
  while (q--) {
    long long m;
    scanf("%lld", &m);
    if (m <= cnt)
      printf("%d %d\n", ans[m][0], ans[m][1]);
    else {
      m -= cnt;
      m %= (n - 1);
      if (!m) m += n - 1;
      printf("%d %d\n", maxVal, who[m]);
    }
  }
  return 0;
}

```