#include <bits/stdc++.h>
using namespace std;
void solve();
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
istream &operator>>(istream &in, vector<T> &ans) {
  for (auto &d : ans) in >> d;
  return in;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &ans) {
  for (auto &d : ans) out << d << " ";
  return out;
}
signed main() {
  fast();
  solve();
}
mt19937_64 rnd(time(0));
struct node {
  int key;
  long long y;
  int sz;
  node *left, *right;
  node(int _key) : key(_key) {
    key = _key;
    y = rnd();
    sz = 1;
    left = nullptr;
    right = nullptr;
  }
};
int size(node *v) { return (v == nullptr) ? 0 : v->sz; }
void upd(node *v) { v->sz = size(v->left) + size(v->right) + 1; }
pair<node *, node *> split_key(node *root, int c) {
  if (root == nullptr) return {nullptr, nullptr};
  if (root->key > c) {
    auto p = split_key(root->left, c);
    root->left = p.second;
    upd(root);
    return {p.first, root};
  } else {
    auto p = split_key(root->right, c);
    root->right = p.first;
    upd(root);
    return {root, p.second};
  }
}
pair<node *, node *> split_size(node *root, int c) {
  if (root == nullptr) return {nullptr, nullptr};
  if (size(root->left) >= c) {
    auto p = split_size(root->left, c);
    root->left = p.second;
    upd(root);
    return {p.first, root};
  } else {
    auto p = split_size(root->right, c - size(root->left) - 1);
    root->right = p.first;
    upd(root);
    return {root, p.second};
  }
}
node *merge(node *a, node *b) {
  if (a == nullptr) return b;
  if (b == nullptr) return a;
  if (a->y > b->y) {
    a->right = merge(a->right, b);
    upd(a);
    return a;
  } else {
    b->left = merge(a, b->left);
    upd(b);
    return b;
  }
}
node *insert(node *root, int ke) {
  auto p = split_key(root, ke);
  return merge(merge(p.first, new node(ke)), p.second);
}
struct qur {
  int k, pos, numb;
};
bool operator<(qur a, qur b) { return a.k < b.k; }
void print(node *root) {
  if (root == nullptr) return;
  print(root->left);
  cout << root->key << " ";
  print(root->right);
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> ans(n), pas;
  for (int i = 0; i < n; ++i) cin >> ans[i].first, ans[i].second = i;
  int m;
  cin >> m;
  vector<qur> kes(m);
  pas = ans;
  for (int i = 0; i < m; ++i) cin >> kes[i].k >> kes[i].pos, kes[i].numb = i;
  auto cmp = [](pair<int, int> a, pair<int, int> b) {
    if (a.first != b.first) return a.first > b.first;
    return a.second < b.second;
  };
  sort(kes.begin(), kes.end());
  sort(ans.begin(), ans.end(), cmp);
  int cnt = 0;
  vector<int> otv(m);
  node *root = nullptr;
  for (auto &[k, pos, numb] : kes) {
    while (cnt < k) root = insert(root, ans[cnt].second), cnt++;
    auto p = split_size(root, pos - 1);
    auto c = split_size(p.second, 1);
    otv[numb] = pas[c.first->key].first;
    root = merge(p.first, merge(c.first, c.second));
  }
  for (auto &d : otv) cout << d << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

This is the harder version of the problem. In this version, 1 ≤ n, m ≤ 2⋅10^5. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems.

You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:

  * [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list); 
  * [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences. 



Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:

  * it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k; 
  * and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal. 



Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:

  * [10, 20, 20] lexicographically less than [10, 21, 1], 
  * [7, 99, 99] is lexicographically less than [10, 21, 1], 
  * [10, 21, 0] is lexicographically less than [10, 21, 1]. 



You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.

For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.

Input

The first line contains an integer n (1 ≤ n ≤ 2⋅10^5) — the length of the sequence a.

The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).

The third line contains an integer m (1 ≤ m ≤ 2⋅10^5) — the number of requests.

The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.

Output

Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.

Examples

Input


3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3


Output


20
10
20
10
20
10


Input


7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4


Output


2
3
2
3
2
3
1
1
3

Note

In the first example, for a=[10,20,10] the optimal subsequences are: 

  * for k=1: [20], 
  * for k=2: [10,20], 
  * for k=3: [10,20,10]. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void solve();
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
istream &operator>>(istream &in, vector<T> &ans) {
  for (auto &d : ans) in >> d;
  return in;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &ans) {
  for (auto &d : ans) out << d << " ";
  return out;
}
signed main() {
  fast();
  solve();
}
mt19937_64 rnd(time(0));
struct node {
  int key;
  long long y;
  int sz;
  node *left, *right;
  node(int _key) : key(_key) {
    key = _key;
    y = rnd();
    sz = 1;
    left = nullptr;
    right = nullptr;
  }
};
int size(node *v) { return (v == nullptr) ? 0 : v->sz; }
void upd(node *v) { v->sz = size(v->left) + size(v->right) + 1; }
pair<node *, node *> split_key(node *root, int c) {
  if (root == nullptr) return {nullptr, nullptr};
  if (root->key > c) {
    auto p = split_key(root->left, c);
    root->left = p.second;
    upd(root);
    return {p.first, root};
  } else {
    auto p = split_key(root->right, c);
    root->right = p.first;
    upd(root);
    return {root, p.second};
  }
}
pair<node *, node *> split_size(node *root, int c) {
  if (root == nullptr) return {nullptr, nullptr};
  if (size(root->left) >= c) {
    auto p = split_size(root->left, c);
    root->left = p.second;
    upd(root);
    return {p.first, root};
  } else {
    auto p = split_size(root->right, c - size(root->left) - 1);
    root->right = p.first;
    upd(root);
    return {root, p.second};
  }
}
node *merge(node *a, node *b) {
  if (a == nullptr) return b;
  if (b == nullptr) return a;
  if (a->y > b->y) {
    a->right = merge(a->right, b);
    upd(a);
    return a;
  } else {
    b->left = merge(a, b->left);
    upd(b);
    return b;
  }
}
node *insert(node *root, int ke) {
  auto p = split_key(root, ke);
  return merge(merge(p.first, new node(ke)), p.second);
}
struct qur {
  int k, pos, numb;
};
bool operator<(qur a, qur b) { return a.k < b.k; }
void print(node *root) {
  if (root == nullptr) return;
  print(root->left);
  cout << root->key << " ";
  print(root->right);
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> ans(n), pas;
  for (int i = 0; i < n; ++i) cin >> ans[i].first, ans[i].second = i;
  int m;
  cin >> m;
  vector<qur> kes(m);
  pas = ans;
  for (int i = 0; i < m; ++i) cin >> kes[i].k >> kes[i].pos, kes[i].numb = i;
  auto cmp = [](pair<int, int> a, pair<int, int> b) {
    if (a.first != b.first) return a.first > b.first;
    return a.second < b.second;
  };
  sort(kes.begin(), kes.end());
  sort(ans.begin(), ans.end(), cmp);
  int cnt = 0;
  vector<int> otv(m);
  node *root = nullptr;
  for (auto &[k, pos, numb] : kes) {
    while (cnt < k) root = insert(root, ans[cnt].second), cnt++;
    auto p = split_size(root, pos - 1);
    auto c = split_size(p.second, 1);
    otv[numb] = pas[c.first->key].first;
    root = merge(p.first, merge(c.first, c.second));
  }
  for (auto &d : otv) cout << d << endl;
}

```