#include <bits/stdc++.h>
#pragma GCC optimize("-Ofast")
using namespace std;
using ll = long long;
ll mod = (ll)1e9 + 7;
ll MOD = mod;
int64_t POW(int64_t a, int64_t b) {
  if (b == 0) return 1;
  a %= mod;
  int64_t res = POW(a, b / 2);
  if (b & 1) return (res % mod * res % mod * a % mod) % mod;
  return (res % mod * res % mod) % mod;
}
ll sum_range2d(ll i, ll j, ll k, ll l, vector<vector<ll>>& sum) {
  return sum[k][l] - sum[k][j - 1] - sum[i - 1][l] + sum[i - 1][j - 1];
}
ll GCDAC(ll a, ll b) {
  if (b == 0) return a;
  return GCDAC(b, a % b);
}
ll gcd(ll a, ll b) {
  if (a < b) swap(a, b);
  return GCDAC(a, b);
}
ll lcm(ll a, ll b) { return (a / gcd(a, b)) * b; }
ll dist(ll X1, ll Y1, ll X2, ll Y2) {
  return sqrt(POW(X1 - X2, 2) + POW(Y1 - Y2, 2));
}
bool intersect(pair<ll, ll> p1, pair<ll, ll> p2) {
  ll x1 = p1.first, x2 = p1.second, y1 = p2.first, y2 = p2.second;
  return (x1 >= y1 && x1 <= y2) || (x2 >= y1 && x2 <= y2) ||
         (y1 >= x1 && y1 <= x2) || (y2 >= x1 && y2 <= x2);
}
bool isPrime(ll n) {
  for (ll i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
ll inverse(ll n) { return POW(n % mod, mod - 2); }
bool mul_overflow(ll x, ll y) { return (log2(x) + log2(y)) >= 64.0; }
ll ceili(ll x, ll y) { return (x + y - 1) / y; }
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  ll x;
  list<ll> lst;
  ll mx = 0;
  for (int i = 0; i < n; i++) {
    cin >> x;
    lst.push_back(x);
    mx = max(mx, x);
  }
  ll a = *lst.begin(), b = *(next(lst.begin()));
  ll qrycnt = 1;
  pair<ll, ll> qryans[2 * n + 3];
  qryans[qrycnt] = make_pair(a, b);
  while (a != mx) {
    if (a > b) {
      lst.erase(next(lst.begin()));
      lst.push_back(b);
    } else {
      lst.erase(lst.begin());
      lst.push_back(a);
    }
    a = *lst.begin(), b = *(next(lst.begin()));
    qryans[++qrycnt] = make_pair(a, b);
  }
  ll arr[n + 3];
  int idx = 1;
  auto it = lst.begin();
  for (; it != lst.end(); ++idx, ++it) {
    arr[idx] = *it;
  }
  while (q--) {
    ll query;
    cin >> query;
    if (query < qrycnt) {
      cout << qryans[query].first << ' ' << qryans[query].second;
    } else {
      cout << mx << ' ' << arr[2 + ((query - qrycnt) % (n - 1))];
    }
    cout << '\n';
  }
}

### 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>
#pragma GCC optimize("-Ofast")
using namespace std;
using ll = long long;
ll mod = (ll)1e9 + 7;
ll MOD = mod;
int64_t POW(int64_t a, int64_t b) {
  if (b == 0) return 1;
  a %= mod;
  int64_t res = POW(a, b / 2);
  if (b & 1) return (res % mod * res % mod * a % mod) % mod;
  return (res % mod * res % mod) % mod;
}
ll sum_range2d(ll i, ll j, ll k, ll l, vector<vector<ll>>& sum) {
  return sum[k][l] - sum[k][j - 1] - sum[i - 1][l] + sum[i - 1][j - 1];
}
ll GCDAC(ll a, ll b) {
  if (b == 0) return a;
  return GCDAC(b, a % b);
}
ll gcd(ll a, ll b) {
  if (a < b) swap(a, b);
  return GCDAC(a, b);
}
ll lcm(ll a, ll b) { return (a / gcd(a, b)) * b; }
ll dist(ll X1, ll Y1, ll X2, ll Y2) {
  return sqrt(POW(X1 - X2, 2) + POW(Y1 - Y2, 2));
}
bool intersect(pair<ll, ll> p1, pair<ll, ll> p2) {
  ll x1 = p1.first, x2 = p1.second, y1 = p2.first, y2 = p2.second;
  return (x1 >= y1 && x1 <= y2) || (x2 >= y1 && x2 <= y2) ||
         (y1 >= x1 && y1 <= x2) || (y2 >= x1 && y2 <= x2);
}
bool isPrime(ll n) {
  for (ll i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
ll inverse(ll n) { return POW(n % mod, mod - 2); }
bool mul_overflow(ll x, ll y) { return (log2(x) + log2(y)) >= 64.0; }
ll ceili(ll x, ll y) { return (x + y - 1) / y; }
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  ll x;
  list<ll> lst;
  ll mx = 0;
  for (int i = 0; i < n; i++) {
    cin >> x;
    lst.push_back(x);
    mx = max(mx, x);
  }
  ll a = *lst.begin(), b = *(next(lst.begin()));
  ll qrycnt = 1;
  pair<ll, ll> qryans[2 * n + 3];
  qryans[qrycnt] = make_pair(a, b);
  while (a != mx) {
    if (a > b) {
      lst.erase(next(lst.begin()));
      lst.push_back(b);
    } else {
      lst.erase(lst.begin());
      lst.push_back(a);
    }
    a = *lst.begin(), b = *(next(lst.begin()));
    qryans[++qrycnt] = make_pair(a, b);
  }
  ll arr[n + 3];
  int idx = 1;
  auto it = lst.begin();
  for (; it != lst.end(); ++idx, ++it) {
    arr[idx] = *it;
  }
  while (q--) {
    ll query;
    cin >> query;
    if (query < qrycnt) {
      cout << qryans[query].first << ' ' << qryans[query].second;
    } else {
      cout << mx << ' ' << arr[2 + ((query - qrycnt) % (n - 1))];
    }
    cout << '\n';
  }
}

```