#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 5;
bool rev(long long x, long long y) { return x > y; }
bool sortinrev(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.first > b.first);
}
long long vec_sum(vector<long long> &v) {
  long long init_sum = 0;
  return accumulate(v.begin(), v.end(), init_sum);
}
long long modulo(long long x, long long N) { return (x % N + N) % N; }
long long max(long long a, long long b) {
  if (a >= b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a <= b)
    return a;
  else
    return b;
}
long long diff(long long a, long long b) {
  if (a >= b)
    return a - b;
  else
    return b - a;
}
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  long long i;
  for (i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long mod = 1e9 + 7;
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
uint64_t mul_mod(uint64_t a, uint64_t b, uint64_t m) {
  long double x;
  uint64_t c;
  int64_t r;
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  x = a;
  c = x * b / m;
  r = (int64_t)(a * b - c * m) % (int64_t)m;
  return r < 0 ? r + m : r;
}
long long n, q;
deque<long long> d;
long long a[N], b[N], ans[N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q;
  cin >> n >> q;
  long long mx = 0;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    mx = max(mx, x);
    d.push_back(x);
  }
  for (long long i = 1; i <= n; i++) {
    a[i] = d.front();
    d.pop_front();
    b[i] = d.front();
    d.pop_front();
    d.push_front(max(a[i], b[i]));
    d.push_back(min(a[i], b[i]));
  }
  d.pop_front();
  for (long long i = 0; i <= n - 2; i++) {
    ans[i] = d.front();
    d.pop_front();
  }
  while (q--) {
    long long x;
    cin >> x;
    if (x <= n) {
      cout << a[x] << " " << b[x] << "\n";
    } else {
      long long idx = x - n;
      idx--;
      idx %= (n - 1);
      cout << mx << " " << ans[idx] << "\n";
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const long long N = 1e5 + 5;
bool rev(long long x, long long y) { return x > y; }
bool sortinrev(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.first > b.first);
}
long long vec_sum(vector<long long> &v) {
  long long init_sum = 0;
  return accumulate(v.begin(), v.end(), init_sum);
}
long long modulo(long long x, long long N) { return (x % N + N) % N; }
long long max(long long a, long long b) {
  if (a >= b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a <= b)
    return a;
  else
    return b;
}
long long diff(long long a, long long b) {
  if (a >= b)
    return a - b;
  else
    return b - a;
}
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  long long i;
  for (i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long mod = 1e9 + 7;
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
uint64_t mul_mod(uint64_t a, uint64_t b, uint64_t m) {
  long double x;
  uint64_t c;
  int64_t r;
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  x = a;
  c = x * b / m;
  r = (int64_t)(a * b - c * m) % (int64_t)m;
  return r < 0 ? r + m : r;
}
long long n, q;
deque<long long> d;
long long a[N], b[N], ans[N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, q;
  cin >> n >> q;
  long long mx = 0;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    mx = max(mx, x);
    d.push_back(x);
  }
  for (long long i = 1; i <= n; i++) {
    a[i] = d.front();
    d.pop_front();
    b[i] = d.front();
    d.pop_front();
    d.push_front(max(a[i], b[i]));
    d.push_back(min(a[i], b[i]));
  }
  d.pop_front();
  for (long long i = 0; i <= n - 2; i++) {
    ans[i] = d.front();
    d.pop_front();
  }
  while (q--) {
    long long x;
    cin >> x;
    if (x <= n) {
      cout << a[x] << " " << b[x] << "\n";
    } else {
      long long idx = x - n;
      idx--;
      idx %= (n - 1);
      cout << mx << " " << ans[idx] << "\n";
    }
  }
  return 0;
}

```