#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
int setBit(int n, int x) { return n | (1 << x); }
int clearBit(int n, int x) { return n & ~(1 << x); }
int toggleBit(int n, int x) { return n ^ (1 << x); }
bool checkBit(int n, int x) { return (bool)(n & (1 << x)); }
int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy8[] = {0, 1, 1, 1, 0, -1, -1, -1};
int dx4[] = {1, 0, -1, 0};
int dy4[] = {0, 1, 0, -1};
const long long inf = 1000000000000000000;
template <typename T>
void amax(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void amin(T& t1, T t2) {
  t1 = min(t1, t2);
}
long long divi = 998244353;
long long __gcd(long long a, long long b) {
  if (a + b == 0) return a + b;
  if (a % b == 0) return b;
  return __gcd(b, a % b);
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) {
    res = res * i;
  }
  return res;
}
long long nCr(long long n, long long r) {
  return fact(n) / (fact(r) * fact(n - r));
}
long long nPr(int n, int r) { return fact(n) / fact(n - r); }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (1LL * res * a) % divi;
    a = (1LL * a * a) % divi;
    b >>= 1;
  }
  return res;
}
void yes() {
  cout << "YES" << endl;
  return;
}
void no() {
  cout << "NO" << endl;
  return;
}
bool isPrime(long long n) {
  if (n == 1) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void prt(vector<int> arr, int n) {
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << endl;
}
void prt(int* arr, int n) {
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << endl;
}
void solve() {
  int n, q;
  cin >> n >> q;
  deque<int> dq;
  int mx = 0;
  int idx = -1;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    dq.push_back(x);
    if (x > mx) {
      mx = x;
      idx = i + 1;
    }
  }
  map<int, pair<int, int> > mp;
  for (int i = 1; i <= idx - 1; i++) {
    int a = dq.front();
    dq.pop_front();
    int b = dq.front();
    dq.pop_front();
    mp[i] = {a, b};
    if (a >= b) {
      dq.push_back(b);
      dq.push_front(a);
    } else {
      dq.push_back(a);
      dq.push_front(b);
    }
  }
  vector<int> v;
  while (dq.size()) {
    int x = dq.front();
    dq.pop_front();
    v.push_back(x);
  }
  for (int i = 0; i < q; i++) {
    long long k;
    cin >> k;
    int a, b;
    if (k < idx) {
      a = mp[k].first;
      b = mp[k].second;
    } else {
      long long rem = k - (idx - 1);
      a = mx;
      long long z = n - 1;
      if (rem % z)
        b = v[rem % z];
      else
        b = v[z];
    }
    cout << a << " " << b << endl;
  }
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

### Prompt

Create a solution in CPP for 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;
using namespace chrono;
int setBit(int n, int x) { return n | (1 << x); }
int clearBit(int n, int x) { return n & ~(1 << x); }
int toggleBit(int n, int x) { return n ^ (1 << x); }
bool checkBit(int n, int x) { return (bool)(n & (1 << x)); }
int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy8[] = {0, 1, 1, 1, 0, -1, -1, -1};
int dx4[] = {1, 0, -1, 0};
int dy4[] = {0, 1, 0, -1};
const long long inf = 1000000000000000000;
template <typename T>
void amax(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void amin(T& t1, T t2) {
  t1 = min(t1, t2);
}
long long divi = 998244353;
long long __gcd(long long a, long long b) {
  if (a + b == 0) return a + b;
  if (a % b == 0) return b;
  return __gcd(b, a % b);
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) {
    res = res * i;
  }
  return res;
}
long long nCr(long long n, long long r) {
  return fact(n) / (fact(r) * fact(n - r));
}
long long nPr(int n, int r) { return fact(n) / fact(n - r); }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (1LL * res * a) % divi;
    a = (1LL * a * a) % divi;
    b >>= 1;
  }
  return res;
}
void yes() {
  cout << "YES" << endl;
  return;
}
void no() {
  cout << "NO" << endl;
  return;
}
bool isPrime(long long n) {
  if (n == 1) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void prt(vector<int> arr, int n) {
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << endl;
}
void prt(int* arr, int n) {
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << endl;
}
void solve() {
  int n, q;
  cin >> n >> q;
  deque<int> dq;
  int mx = 0;
  int idx = -1;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    dq.push_back(x);
    if (x > mx) {
      mx = x;
      idx = i + 1;
    }
  }
  map<int, pair<int, int> > mp;
  for (int i = 1; i <= idx - 1; i++) {
    int a = dq.front();
    dq.pop_front();
    int b = dq.front();
    dq.pop_front();
    mp[i] = {a, b};
    if (a >= b) {
      dq.push_back(b);
      dq.push_front(a);
    } else {
      dq.push_back(a);
      dq.push_front(b);
    }
  }
  vector<int> v;
  while (dq.size()) {
    int x = dq.front();
    dq.pop_front();
    v.push_back(x);
  }
  for (int i = 0; i < q; i++) {
    long long k;
    cin >> k;
    int a, b;
    if (k < idx) {
      a = mp[k].first;
      b = mp[k].second;
    } else {
      long long rem = k - (idx - 1);
      a = mx;
      long long z = n - 1;
      if (rem % z)
        b = v[rem % z];
      else
        b = v[z];
    }
    cout << a << " " << b << endl;
  }
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

```