#include <bits/stdc++.h>
using namespace std;
long long powmod(long long a, long long b, long long mod) {
  if (b == 0 || a == 1) return mod == 1 ? 0 : 1;
  if (b % 2 == 0) {
    long long k = powmod(a, b / 2, mod);
    return (k * k) % mod;
  } else {
    long long k = powmod(a, b / 2, mod);
    return ((k * k) % mod * a) % mod;
  }
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
bool is_prime(long long n) {
  long long bound = sqrt(n);
  long long d = 2;
  while (d <= bound && n % d != 0) ++d;
  return d * d > n;
}
const int MOD = 1000000007;
const int INF = 1000000050;
const int MX = 100010;
const double EPS = 1e-9;
mt19937 rnd;
struct Node {
  int prior, key, sz;
  Node *l, *r;
  Node(int key) : key(key), prior(rnd()), sz(1), l(nullptr), r(nullptr){};
};
int GetSize(Node* t) {
  if (t == nullptr) {
    return 0;
  } else {
    return t->sz;
  }
}
void Update(Node* t) { t->sz = GetSize(t->l) + GetSize(t->r) + 1; }
pair<Node*, Node*> Split(Node* t, int key) {
  if (t == nullptr) {
    return {nullptr, nullptr};
  }
  if (t->key < key) {
    auto splited = Split(t->r, key);
    t->r = splited.first;
    Update(t);
    return {t, splited.second};
  } else {
    auto splited = Split(t->l, key);
    t->l = splited.second;
    Update(t);
    return {splited.first, t};
  }
}
Node* Merge(Node* t1, Node* t2) {
  if (t1 == nullptr) {
    return t2;
  }
  if (t2 == nullptr) {
    return t1;
  }
  if (t1->prior > t2->prior) {
    t1->r = Merge(t1->r, t2);
    Update(t1);
    return t1;
  } else {
    t2->l = Merge(t1, t2->l);
    Update(t2);
    return t2;
  }
}
Node* Insert(Node* t, int key) {
  Node* key_node = new Node(key);
  auto splited_left = Split(t, key);
  auto splited_right = Split(splited_left.second, key + 1);
  t = Merge(splited_left.first, key_node);
  t = Merge(t, splited_right.second);
  Update(t);
  return t;
}
int FindKth(Node* t, int k) {
  if (t == nullptr) {
    return INF;
  }
  if (GetSize(t->l) + 1 == k) {
    return t->key;
  } else if (GetSize(t->l) >= k) {
    return FindKth(t->l, k);
  } else {
    return FindKth(t->r, k - 1 - GetSize(t->l));
  }
}
map<int, set<int>> ps;
struct req {
  int k;
  int pos;
  int id;
};
bool cmp(req a, req b) { return a.k < b.k; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> q(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ps[a[i]].insert(i);
    q[i] = a[i];
  }
  sort(begin(q), end(q));
  reverse(begin(q), end(q));
  vector<int> arr;
  int pr = -1;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    int pos = *(ps[q[i]].begin());
    arr.push_back(pos);
    ps[q[i]].erase(ps[q[i]].begin());
  }
  int m;
  cin >> m;
  vector<int> ans(m);
  vector<req> reqs(m);
  for (int i = 0; i < m; i++) {
    int pos, k;
    cin >> k >> pos;
    reqs[i] = {k, pos, i};
  }
  sort(begin(reqs), end(reqs), cmp);
  int prv = 0;
  Node* t = nullptr;
  for (req r : reqs) {
    int k = r.k;
    int pos = r.pos;
    for (int i = prv; i < k; i++) {
      t = Insert(t, arr[i]);
    }
    prv = k;
    int id = r.id;
    ans[id] = FindKth(t, pos);
  }
  for (int i = 0; i < m; i++) cout << a[ans[i]] << '\n';
}

### Prompt

In Cpp, your task is to solve the following problem:

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;
long long powmod(long long a, long long b, long long mod) {
  if (b == 0 || a == 1) return mod == 1 ? 0 : 1;
  if (b % 2 == 0) {
    long long k = powmod(a, b / 2, mod);
    return (k * k) % mod;
  } else {
    long long k = powmod(a, b / 2, mod);
    return ((k * k) % mod * a) % mod;
  }
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
bool is_prime(long long n) {
  long long bound = sqrt(n);
  long long d = 2;
  while (d <= bound && n % d != 0) ++d;
  return d * d > n;
}
const int MOD = 1000000007;
const int INF = 1000000050;
const int MX = 100010;
const double EPS = 1e-9;
mt19937 rnd;
struct Node {
  int prior, key, sz;
  Node *l, *r;
  Node(int key) : key(key), prior(rnd()), sz(1), l(nullptr), r(nullptr){};
};
int GetSize(Node* t) {
  if (t == nullptr) {
    return 0;
  } else {
    return t->sz;
  }
}
void Update(Node* t) { t->sz = GetSize(t->l) + GetSize(t->r) + 1; }
pair<Node*, Node*> Split(Node* t, int key) {
  if (t == nullptr) {
    return {nullptr, nullptr};
  }
  if (t->key < key) {
    auto splited = Split(t->r, key);
    t->r = splited.first;
    Update(t);
    return {t, splited.second};
  } else {
    auto splited = Split(t->l, key);
    t->l = splited.second;
    Update(t);
    return {splited.first, t};
  }
}
Node* Merge(Node* t1, Node* t2) {
  if (t1 == nullptr) {
    return t2;
  }
  if (t2 == nullptr) {
    return t1;
  }
  if (t1->prior > t2->prior) {
    t1->r = Merge(t1->r, t2);
    Update(t1);
    return t1;
  } else {
    t2->l = Merge(t1, t2->l);
    Update(t2);
    return t2;
  }
}
Node* Insert(Node* t, int key) {
  Node* key_node = new Node(key);
  auto splited_left = Split(t, key);
  auto splited_right = Split(splited_left.second, key + 1);
  t = Merge(splited_left.first, key_node);
  t = Merge(t, splited_right.second);
  Update(t);
  return t;
}
int FindKth(Node* t, int k) {
  if (t == nullptr) {
    return INF;
  }
  if (GetSize(t->l) + 1 == k) {
    return t->key;
  } else if (GetSize(t->l) >= k) {
    return FindKth(t->l, k);
  } else {
    return FindKth(t->r, k - 1 - GetSize(t->l));
  }
}
map<int, set<int>> ps;
struct req {
  int k;
  int pos;
  int id;
};
bool cmp(req a, req b) { return a.k < b.k; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> q(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ps[a[i]].insert(i);
    q[i] = a[i];
  }
  sort(begin(q), end(q));
  reverse(begin(q), end(q));
  vector<int> arr;
  int pr = -1;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    int pos = *(ps[q[i]].begin());
    arr.push_back(pos);
    ps[q[i]].erase(ps[q[i]].begin());
  }
  int m;
  cin >> m;
  vector<int> ans(m);
  vector<req> reqs(m);
  for (int i = 0; i < m; i++) {
    int pos, k;
    cin >> k >> pos;
    reqs[i] = {k, pos, i};
  }
  sort(begin(reqs), end(reqs), cmp);
  int prv = 0;
  Node* t = nullptr;
  for (req r : reqs) {
    int k = r.k;
    int pos = r.pos;
    for (int i = prv; i < k; i++) {
      t = Insert(t, arr[i]);
    }
    prv = k;
    int id = r.id;
    ans[id] = FindKth(t, pos);
  }
  for (int i = 0; i < m; i++) cout << a[ans[i]] << '\n';
}

```