#include <bits/stdc++.h>
using namespace std;
long long MOD = 1e18 + 7;
long double EPS = 1e-12;
struct Node {
  long long x;
  Node *l, *r;
  Node(long long a, Node *left, Node *right) {
    x = a;
    l = left;
    r = right;
  }
};
Node *build(long long l, long long r) {
  if (l + 1 == r) {
    return new Node(0, nullptr, nullptr);
  }
  long long m = (l + r) / 2;
  Node *u = build(l, m), *v = build(m, r);
  return new Node(0, u, v);
}
Node *update(long long l, long long r, long long q, Node *root) {
  if (l > q || r <= q) return root;
  if (l + 1 == r) {
    return new Node(root->x + 1, nullptr, nullptr);
  }
  long long m = (l + r) / 2;
  Node *u = update(l, m, q, root->l), *v = update(m, r, q, root->r);
  return new Node(u->x + v->x, u, v);
}
long long get(long long l, long long r, long long ql, long long qr,
              Node *root) {
  if (l >= qr || r <= ql) return 0;
  if (l >= ql && r <= qr) return root->x;
  long long m = (l + r) / 2;
  return get(l, m, ql, qr, root->l) + get(m, r, ql, qr, root->r);
}
bool lesss(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
signed main() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (auto &i : arr) cin >> i;
  long long m;
  cin >> m;
  vector<pair<long long, long long>> tosort;
  for (long long i = 0; i < n; i++) {
    tosort.push_back({arr[i], i});
  }
  sort(tosort.begin(), tosort.end(), lesss);
  vector<Node *> nodes;
  Node *tree = build(0, n);
  for (long long i = 0; i < n; i++) {
    if (i == 0) {
      nodes.push_back(update(0, n, tosort[i].second, tree));
    } else {
      nodes.push_back(update(0, n, tosort[i].second, nodes[i - 1]));
    }
  }
  for (long long i = 0; i < m; i++) {
    long long k, pos;
    cin >> k >> pos;
    long long l = 0, r = n + 1;
    while (l + 1 != r) {
      long long mm = (l + r) / 2;
      if (get(0, n, 0, mm, nodes[k - 1]) >= pos)
        r = mm;
      else
        l = mm;
    }
    cout << arr[r - 1] << endl;
  }
}

### 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;
long long MOD = 1e18 + 7;
long double EPS = 1e-12;
struct Node {
  long long x;
  Node *l, *r;
  Node(long long a, Node *left, Node *right) {
    x = a;
    l = left;
    r = right;
  }
};
Node *build(long long l, long long r) {
  if (l + 1 == r) {
    return new Node(0, nullptr, nullptr);
  }
  long long m = (l + r) / 2;
  Node *u = build(l, m), *v = build(m, r);
  return new Node(0, u, v);
}
Node *update(long long l, long long r, long long q, Node *root) {
  if (l > q || r <= q) return root;
  if (l + 1 == r) {
    return new Node(root->x + 1, nullptr, nullptr);
  }
  long long m = (l + r) / 2;
  Node *u = update(l, m, q, root->l), *v = update(m, r, q, root->r);
  return new Node(u->x + v->x, u, v);
}
long long get(long long l, long long r, long long ql, long long qr,
              Node *root) {
  if (l >= qr || r <= ql) return 0;
  if (l >= ql && r <= qr) return root->x;
  long long m = (l + r) / 2;
  return get(l, m, ql, qr, root->l) + get(m, r, ql, qr, root->r);
}
bool lesss(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
signed main() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (auto &i : arr) cin >> i;
  long long m;
  cin >> m;
  vector<pair<long long, long long>> tosort;
  for (long long i = 0; i < n; i++) {
    tosort.push_back({arr[i], i});
  }
  sort(tosort.begin(), tosort.end(), lesss);
  vector<Node *> nodes;
  Node *tree = build(0, n);
  for (long long i = 0; i < n; i++) {
    if (i == 0) {
      nodes.push_back(update(0, n, tosort[i].second, tree));
    } else {
      nodes.push_back(update(0, n, tosort[i].second, nodes[i - 1]));
    }
  }
  for (long long i = 0; i < m; i++) {
    long long k, pos;
    cin >> k >> pos;
    long long l = 0, r = n + 1;
    while (l + 1 != r) {
      long long mm = (l + r) / 2;
      if (get(0, n, 0, mm, nodes[k - 1]) >= pos)
        r = mm;
      else
        l = mm;
    }
    cout << arr[r - 1] << endl;
  }
}

```