#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e12;
long long SEED = 1e6 + 3;
long long mod = 1e9;
struct Node {
  long long x, y, s;
  Node *left, *right;
  Node() {}
  Node(long long val) {
    s = 1;
    x = val;
    y = rand();
    left = nullptr;
    right = nullptr;
  }
  ~Node() {
    if (left) delete left;
    if (right) delete right;
  }
};
long long get_size(Node *root) {
  if (root) return root->s;
  return 0;
}
void update(Node *root) {
  if (!root) return;
  root->s = 1 + get_size(root->left) + get_size(root->right);
}
bool find_e(Node *root, long long key) {
  if (!root) return false;
  if (root->x == key)
    return true;
  else if (root->x > key)
    return find_e(root->left, key);
  else
    return find_e(root->right, key);
}
pair<Node *, Node *> split(Node *root, long long key) {
  if (!root) return {nullptr, nullptr};
  if (root->x <= key) {
    pair<Node *, Node *> roots;
    roots = split(root->right, key);
    root->right = roots.first;
    update(root);
    return {root, roots.second};
  } else {
    pair<Node *, Node *> roots;
    roots = split(root->left, key);
    root->left = roots.second;
    update(root);
    return {roots.first, root};
  }
}
Node *merge_e(Node *root1, Node *root2) {
  if (!root1) return root2;
  if (!root2) return root1;
  if (root1->y > root2->y) {
    root1->right = merge_e(root1->right, root2);
    update(root1);
    return root1;
  } else {
    root2->left = merge_e(root1, root2->left);
    update(root2);
    return root2;
  }
}
Node *insert_e(Node *root, long long key) {
  if (find_e(root, key)) return root;
  pair<Node *, Node *> roots = split(root, key);
  Node *root2 = new Node(key);
  return merge_e(merge_e(roots.first, root2), roots.second);
}
long long get_k(Node *root, long long k) {
  long long s1 = get_size(root->left);
  if (k == s1)
    return root->x;
  else if (k < s1)
    return get_k(root->left, k);
  else
    return get_k(root->right, k - s1 - 1);
}
int32_t main() {
  srand(SEED);
  long long n;
  cin >> n;
  vector<long long> a(n);
  vector<vector<long long>> pr(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long m;
  cin >> m;
  vector<vector<long long>> q(m, vector<long long>(3));
  vector<vector<long long>> l(n, vector<long long>(2));
  for (long long i = 0; i < m; i++) {
    cin >> q[i][0] >> q[i][1];
    q[i][2] = i;
  }
  sort(q.begin(), q.end());
  for (long long i = 0; i < n; i++) {
    l[i][0] = a[i];
    l[i][1] = -i;
  }
  sort(l.begin(), l.end());
  reverse(l.begin(), l.end());
  vector<long long> ans(m);
  long long j = 0;
  Node *root = nullptr;
  for (long long i = 0; i < m; i++) {
    long long now_k = q[i][0], now_pos = q[i][1];
    now_k--;
    now_pos--;
    while (j <= now_k) {
      root = insert_e(root, -l[j][1]);
      j++;
    }
    long long lans = get_k(root, now_pos);
    ans[q[i][2]] = a[lans];
  }
  for (auto v : ans) cout << v << "\n";
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const long long INF = 1e12;
long long SEED = 1e6 + 3;
long long mod = 1e9;
struct Node {
  long long x, y, s;
  Node *left, *right;
  Node() {}
  Node(long long val) {
    s = 1;
    x = val;
    y = rand();
    left = nullptr;
    right = nullptr;
  }
  ~Node() {
    if (left) delete left;
    if (right) delete right;
  }
};
long long get_size(Node *root) {
  if (root) return root->s;
  return 0;
}
void update(Node *root) {
  if (!root) return;
  root->s = 1 + get_size(root->left) + get_size(root->right);
}
bool find_e(Node *root, long long key) {
  if (!root) return false;
  if (root->x == key)
    return true;
  else if (root->x > key)
    return find_e(root->left, key);
  else
    return find_e(root->right, key);
}
pair<Node *, Node *> split(Node *root, long long key) {
  if (!root) return {nullptr, nullptr};
  if (root->x <= key) {
    pair<Node *, Node *> roots;
    roots = split(root->right, key);
    root->right = roots.first;
    update(root);
    return {root, roots.second};
  } else {
    pair<Node *, Node *> roots;
    roots = split(root->left, key);
    root->left = roots.second;
    update(root);
    return {roots.first, root};
  }
}
Node *merge_e(Node *root1, Node *root2) {
  if (!root1) return root2;
  if (!root2) return root1;
  if (root1->y > root2->y) {
    root1->right = merge_e(root1->right, root2);
    update(root1);
    return root1;
  } else {
    root2->left = merge_e(root1, root2->left);
    update(root2);
    return root2;
  }
}
Node *insert_e(Node *root, long long key) {
  if (find_e(root, key)) return root;
  pair<Node *, Node *> roots = split(root, key);
  Node *root2 = new Node(key);
  return merge_e(merge_e(roots.first, root2), roots.second);
}
long long get_k(Node *root, long long k) {
  long long s1 = get_size(root->left);
  if (k == s1)
    return root->x;
  else if (k < s1)
    return get_k(root->left, k);
  else
    return get_k(root->right, k - s1 - 1);
}
int32_t main() {
  srand(SEED);
  long long n;
  cin >> n;
  vector<long long> a(n);
  vector<vector<long long>> pr(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long m;
  cin >> m;
  vector<vector<long long>> q(m, vector<long long>(3));
  vector<vector<long long>> l(n, vector<long long>(2));
  for (long long i = 0; i < m; i++) {
    cin >> q[i][0] >> q[i][1];
    q[i][2] = i;
  }
  sort(q.begin(), q.end());
  for (long long i = 0; i < n; i++) {
    l[i][0] = a[i];
    l[i][1] = -i;
  }
  sort(l.begin(), l.end());
  reverse(l.begin(), l.end());
  vector<long long> ans(m);
  long long j = 0;
  Node *root = nullptr;
  for (long long i = 0; i < m; i++) {
    long long now_k = q[i][0], now_pos = q[i][1];
    now_k--;
    now_pos--;
    while (j <= now_k) {
      root = insert_e(root, -l[j][1]);
      j++;
    }
    long long lans = get_k(root, now_pos);
    ans[q[i][2]] = a[lans];
  }
  for (auto v : ans) cout << v << "\n";
  return 0;
}

```