#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
long long int shouldendl = 1;
long long int MOD = (long long int)(1e9 + 7);
using namespace std;
long long int modInverse(long long int a) {
  long long int m = MOD, m0 = m, y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m, t = m;
    m = a % m, a = t, t = y;
    y = x - q * y, x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long int add(long long a, long long b) {
  return ((a % MOD) + (b % MOD)) % MOD;
}
long long int mul(long long a, long long b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
long long int sub(long long a, long long b) {
  long long int ans = ((a % MOD - b % MOD)) % MOD;
  if (ans < 0) ans += MOD;
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void MAIN(long long int i, long long int t);
int get() {
  long long int t;
  cin >> t;
  ;
  return t;
}
template <class T>
int logd(T name) {
  cout << name << endl;
  return 0;
}
void test(int t) {
  for (long long int i = 1; i <= t; i++) {
    MAIN(i, t);
    if (shouldendl) cout << endl;
  }
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % MOD;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1;
    x = (x * x) % MOD;
  }
  return res;
}
template <typename T, typename U>
std::pair<T, U> operator+(const std::pair<T, U> &l, const std::pair<T, U> &r) {
  return {l.first + r.first, l.second + r.second};
}
long long int popcount(long long int x) { return __builtin_popcountll(x); }
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
template <class T>
bool input(vector<T> &v, long long int n = -1) {
  assert(v.size() != 0);
  if (n == -1) n = v.size();
  for (auto &i : v) cin >> i;
  return 0;
}
template <class T>
bool print(vector<T> v, long long int l = 0, long long int r = -1) {
  if (r == -1) r = v.size();
  for (int i = l; i < r; i++) cout << v[i] << " ";
  return 0;
}
void MAIN(long long int current_test_case, long long int total_test_cases) {
  vector<pair<long long int, long long int> > temp;
  long long int n, m;
  cin >> n >> m;
  deque<long long int> q;
  long long int max1 = 0;
  vector<long long int> v;
  for (long long int i = 0; i < n; i++) {
    long long int temp;
    cin >> temp;
    max1 = max(temp, max1);
    q.push_back(temp);
  }
  for (long long int i = 0; i < n; i++) {
    long long int first = q.front();
    if (first == max1) break;
    q.pop_front();
    long long int second = q.front();
    q.pop_front();
    temp.push_back(make_pair(first, second));
    if (first > second)
      q.push_front(first), q.push_back(second);
    else
      q.push_front(second), q.push_back(first);
  }
  while (!q.empty()) {
    v.push_back(q.front());
    q.pop_front();
  }
  vector<pair<long long int, long long int> > extra;
  for (int i = 1; i < n; i++) {
    extra.push_back(make_pair(v[0], v[i]));
  }
  while (m--) {
    long long int index;
    cin >> index;
    if (index <= temp.size())
      cout << temp[index - 1].first << " " << temp[index - 1].second;
    else {
      int check = extra.size();
      index = index - temp.size() - 1;
      cout << extra[index % check].first << " " << extra[index % check].second;
    }
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  test(1);
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
long long int shouldendl = 1;
long long int MOD = (long long int)(1e9 + 7);
using namespace std;
long long int modInverse(long long int a) {
  long long int m = MOD, m0 = m, y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m, t = m;
    m = a % m, a = t, t = y;
    y = x - q * y, x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long int add(long long a, long long b) {
  return ((a % MOD) + (b % MOD)) % MOD;
}
long long int mul(long long a, long long b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
long long int sub(long long a, long long b) {
  long long int ans = ((a % MOD - b % MOD)) % MOD;
  if (ans < 0) ans += MOD;
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void MAIN(long long int i, long long int t);
int get() {
  long long int t;
  cin >> t;
  ;
  return t;
}
template <class T>
int logd(T name) {
  cout << name << endl;
  return 0;
}
void test(int t) {
  for (long long int i = 1; i <= t; i++) {
    MAIN(i, t);
    if (shouldendl) cout << endl;
  }
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % MOD;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1;
    x = (x * x) % MOD;
  }
  return res;
}
template <typename T, typename U>
std::pair<T, U> operator+(const std::pair<T, U> &l, const std::pair<T, U> &r) {
  return {l.first + r.first, l.second + r.second};
}
long long int popcount(long long int x) { return __builtin_popcountll(x); }
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
template <class T>
bool input(vector<T> &v, long long int n = -1) {
  assert(v.size() != 0);
  if (n == -1) n = v.size();
  for (auto &i : v) cin >> i;
  return 0;
}
template <class T>
bool print(vector<T> v, long long int l = 0, long long int r = -1) {
  if (r == -1) r = v.size();
  for (int i = l; i < r; i++) cout << v[i] << " ";
  return 0;
}
void MAIN(long long int current_test_case, long long int total_test_cases) {
  vector<pair<long long int, long long int> > temp;
  long long int n, m;
  cin >> n >> m;
  deque<long long int> q;
  long long int max1 = 0;
  vector<long long int> v;
  for (long long int i = 0; i < n; i++) {
    long long int temp;
    cin >> temp;
    max1 = max(temp, max1);
    q.push_back(temp);
  }
  for (long long int i = 0; i < n; i++) {
    long long int first = q.front();
    if (first == max1) break;
    q.pop_front();
    long long int second = q.front();
    q.pop_front();
    temp.push_back(make_pair(first, second));
    if (first > second)
      q.push_front(first), q.push_back(second);
    else
      q.push_front(second), q.push_back(first);
  }
  while (!q.empty()) {
    v.push_back(q.front());
    q.pop_front();
  }
  vector<pair<long long int, long long int> > extra;
  for (int i = 1; i < n; i++) {
    extra.push_back(make_pair(v[0], v[i]));
  }
  while (m--) {
    long long int index;
    cin >> index;
    if (index <= temp.size())
      cout << temp[index - 1].first << " " << temp[index - 1].second;
    else {
      int check = extra.size();
      index = index - temp.size() - 1;
      cout << extra[index % check].first << " " << extra[index % check].second;
    }
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  test(1);
  return 0;
}

```