#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9;
bool cmp(pair<int, int> a, pair<int, int> b) {
  return a.first > b.first || (a.first == b.first && a.second < b.second);
}
struct Node {
  pair<int, int> x;
  int y;
  Node* l;
  Node* r;
  int sz;
  Node(pair<int, int> x) : x(x), y(rand()), sz(1), l(nullptr), r(nullptr){};
};
int getsz(Node* t) {
  if (t == nullptr) {
    return 0;
  }
  return t->sz;
}
Node* update(Node* t) {
  t->sz = getsz(t->l) + getsz(t->r) + 1;
  return t;
}
Node* merge(Node* l, Node* r) {
  if (l == nullptr) return r;
  if (r == nullptr) return l;
  if (l->y > r->y) {
    l->r = merge(l->r, r);
    return update(l);
  } else {
    r->l = merge(l, r->l);
    return update(r);
  }
}
pair<Node*, Node*> split(Node* t, pair<int, int> x) {
  pair<Node*, Node*> ptt = {nullptr, nullptr};
  if (t == nullptr) {
    return ptt;
  }
  if (x <= t->x) {
    ptt = split(t->l, x);
    t->l = ptt.second;
    update(t);
    ptt.second = t;
    return {ptt.first, t};
  } else {
    ptt = split(t->r, x);
    t->r = ptt.first;
    update(t);
    ptt.first = t;
    return {t, ptt.second};
  }
  return ptt;
}
pair<Node*, Node*> split_ind(Node* t, int x) {
  pair<Node*, Node*> ptt = {nullptr, nullptr};
  if (t == nullptr) {
    return ptt;
  }
  if (getsz(t->l) < x) {
    ptt = split_ind(t->r, x - getsz(t->l) - 1);
    t->r = ptt.first;
    ptt.first = t;
  } else {
    ptt = split_ind(t->l, x);
    t->l = ptt.second;
    ptt.second = t;
  }
  update(t);
  return ptt;
}
pair<int, int> get(Node* t, int x) {
  pair<Node*, Node*> first = split_ind(t, x);
  pair<Node*, Node*> second = split_ind(first.first, x - 1);
  pair<int, int> result = second.second->x;
  t = merge(merge(second.first, second.second), first.second);
  return result;
}
Node* insert(Node* t, pair<int, int> x) {
  pair<Node*, Node*> res = split(t, x);
  Node* n = new Node(x);
  if (res.first != nullptr && n->x < res.first->x) {
    t = merge(merge(n, res.first), res.second);
    return t;
  }
  if (res.second != nullptr && n->x < res.second->x) {
    t = merge(merge(res.first, n), res.second);
    return t;
  }
  t = merge(t, n);
  return t;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a.begin(), a.end(), cmp);
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int>> q(m);
  for (int i = 0; i < m; i++) {
    cin >> q[i].first.first >> q[i].first.second;
    q[i].second = i;
  }
  sort(q.begin(), q.end());
  int pk = 0;
  Node* ROOT = nullptr;
  vector<int> ans(m);
  for (int i = 0; i < m; i++) {
    int k = q[i].first.first;
    int pos = q[i].first.second;
    for (int j = pk; j < k; j++) {
      ROOT = insert(ROOT, {a[j].second, a[j].first});
    }
    pair<int, int> res = get(ROOT, pos);
    ans[q[i].second] = res.second;
    pk = k;
  }
  for (auto i : ans) {
    cout << i << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const long long INF = 1e9;
bool cmp(pair<int, int> a, pair<int, int> b) {
  return a.first > b.first || (a.first == b.first && a.second < b.second);
}
struct Node {
  pair<int, int> x;
  int y;
  Node* l;
  Node* r;
  int sz;
  Node(pair<int, int> x) : x(x), y(rand()), sz(1), l(nullptr), r(nullptr){};
};
int getsz(Node* t) {
  if (t == nullptr) {
    return 0;
  }
  return t->sz;
}
Node* update(Node* t) {
  t->sz = getsz(t->l) + getsz(t->r) + 1;
  return t;
}
Node* merge(Node* l, Node* r) {
  if (l == nullptr) return r;
  if (r == nullptr) return l;
  if (l->y > r->y) {
    l->r = merge(l->r, r);
    return update(l);
  } else {
    r->l = merge(l, r->l);
    return update(r);
  }
}
pair<Node*, Node*> split(Node* t, pair<int, int> x) {
  pair<Node*, Node*> ptt = {nullptr, nullptr};
  if (t == nullptr) {
    return ptt;
  }
  if (x <= t->x) {
    ptt = split(t->l, x);
    t->l = ptt.second;
    update(t);
    ptt.second = t;
    return {ptt.first, t};
  } else {
    ptt = split(t->r, x);
    t->r = ptt.first;
    update(t);
    ptt.first = t;
    return {t, ptt.second};
  }
  return ptt;
}
pair<Node*, Node*> split_ind(Node* t, int x) {
  pair<Node*, Node*> ptt = {nullptr, nullptr};
  if (t == nullptr) {
    return ptt;
  }
  if (getsz(t->l) < x) {
    ptt = split_ind(t->r, x - getsz(t->l) - 1);
    t->r = ptt.first;
    ptt.first = t;
  } else {
    ptt = split_ind(t->l, x);
    t->l = ptt.second;
    ptt.second = t;
  }
  update(t);
  return ptt;
}
pair<int, int> get(Node* t, int x) {
  pair<Node*, Node*> first = split_ind(t, x);
  pair<Node*, Node*> second = split_ind(first.first, x - 1);
  pair<int, int> result = second.second->x;
  t = merge(merge(second.first, second.second), first.second);
  return result;
}
Node* insert(Node* t, pair<int, int> x) {
  pair<Node*, Node*> res = split(t, x);
  Node* n = new Node(x);
  if (res.first != nullptr && n->x < res.first->x) {
    t = merge(merge(n, res.first), res.second);
    return t;
  }
  if (res.second != nullptr && n->x < res.second->x) {
    t = merge(merge(res.first, n), res.second);
    return t;
  }
  t = merge(t, n);
  return t;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a.begin(), a.end(), cmp);
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int>> q(m);
  for (int i = 0; i < m; i++) {
    cin >> q[i].first.first >> q[i].first.second;
    q[i].second = i;
  }
  sort(q.begin(), q.end());
  int pk = 0;
  Node* ROOT = nullptr;
  vector<int> ans(m);
  for (int i = 0; i < m; i++) {
    int k = q[i].first.first;
    int pos = q[i].first.second;
    for (int j = pk; j < k; j++) {
      ROOT = insert(ROOT, {a[j].second, a[j].first});
    }
    pair<int, int> res = get(ROOT, pos);
    ans[q[i].second] = res.second;
    pk = k;
  }
  for (auto i : ans) {
    cout << i << '\n';
  }
  return 0;
}

```