#include <bits/stdc++.h>
using namespace std;
vector<int> t;
void update(int v, int lv, int rv, int pos) {
  if (lv == rv) {
    t[v] = 1;
    return;
  }
  int m = (lv + rv) >> 1;
  if (pos <= m)
    update((v << 1) + 1, lv, m, pos);
  else
    update((v << 1) + 2, m + 1, rv, pos);
  t[v] = t[(v << 1) + 1] + t[(v << 1) + 2];
}
int get(int v, int lv, int rv, int l, int r) {
  if (lv == l && rv == r) return t[v];
  int m = (lv + rv) >> 1;
  if (r <= m) return get((v << 1) + 1, lv, m, l, r);
  if (l > m)
    return get((v << 1) + 2, m + 1, rv, l, r);
  else
    return get((v << 1) + 1, lv, m, l, m) +
           get((v << 1) + 2, m + 1, rv, m + 1, r);
}
void print(int v, int lv, int rv) {
  if (lv == rv) {
    cout << t[v] << ' ';
    return;
  }
  int m = (lv + rv) >> 1;
  print((v << 1) + 1, lv, m);
  print((v << 1) + 2, m + 1, rv);
}
int main() {
  int n;
  cin >> n;
  int a[n];
  pair<int, int> order[n];
  for (int i = 0; i < n; ++i) {
    cin >> order[i].first;
    order[i].second = i;
    a[i] = order[i].first;
  }
  sort(order, order + n, [](pair<int, int>& l, pair<int, int>& r) {
    if (l.first != r.first)
      return l.first > r.first;
    else
      return l.second < r.second;
  });
  int T;
  cin >> T;
  pair<pair<int, int>, int> query[T];
  for (int i = 0; i < T; ++i) {
    cin >> query[i].first.first >> query[i].first.second;
    query[i].second = i;
  }
  sort(query, query + T);
  t.resize(4 * n, 0);
  int lk = -1;
  int answer[T];
  for (auto q : query) {
    int k = q.first.first, pos = q.first.second;
    for (int i = lk + 1; i < k; ++i) {
      update(0, 0, n - 1, order[i].second);
    }
    int l = -1, r = n - 1;
    while (l + 1 != r) {
      int m = (l + r) >> 1;
      if (get(0, 0, n - 1, 0, m) < pos)
        l = m;
      else
        r = m;
    }
    answer[q.second] = a[r];
    lk = k - 1;
  }
  for (int x : answer) {
    cout << x << ' ';
  }
  return 0;
}

### 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;
vector<int> t;
void update(int v, int lv, int rv, int pos) {
  if (lv == rv) {
    t[v] = 1;
    return;
  }
  int m = (lv + rv) >> 1;
  if (pos <= m)
    update((v << 1) + 1, lv, m, pos);
  else
    update((v << 1) + 2, m + 1, rv, pos);
  t[v] = t[(v << 1) + 1] + t[(v << 1) + 2];
}
int get(int v, int lv, int rv, int l, int r) {
  if (lv == l && rv == r) return t[v];
  int m = (lv + rv) >> 1;
  if (r <= m) return get((v << 1) + 1, lv, m, l, r);
  if (l > m)
    return get((v << 1) + 2, m + 1, rv, l, r);
  else
    return get((v << 1) + 1, lv, m, l, m) +
           get((v << 1) + 2, m + 1, rv, m + 1, r);
}
void print(int v, int lv, int rv) {
  if (lv == rv) {
    cout << t[v] << ' ';
    return;
  }
  int m = (lv + rv) >> 1;
  print((v << 1) + 1, lv, m);
  print((v << 1) + 2, m + 1, rv);
}
int main() {
  int n;
  cin >> n;
  int a[n];
  pair<int, int> order[n];
  for (int i = 0; i < n; ++i) {
    cin >> order[i].first;
    order[i].second = i;
    a[i] = order[i].first;
  }
  sort(order, order + n, [](pair<int, int>& l, pair<int, int>& r) {
    if (l.first != r.first)
      return l.first > r.first;
    else
      return l.second < r.second;
  });
  int T;
  cin >> T;
  pair<pair<int, int>, int> query[T];
  for (int i = 0; i < T; ++i) {
    cin >> query[i].first.first >> query[i].first.second;
    query[i].second = i;
  }
  sort(query, query + T);
  t.resize(4 * n, 0);
  int lk = -1;
  int answer[T];
  for (auto q : query) {
    int k = q.first.first, pos = q.first.second;
    for (int i = lk + 1; i < k; ++i) {
      update(0, 0, n - 1, order[i].second);
    }
    int l = -1, r = n - 1;
    while (l + 1 != r) {
      int m = (l + r) >> 1;
      if (get(0, 0, n - 1, 0, m) < pos)
        l = m;
      else
        r = m;
    }
    answer[q.second] = a[r];
    lk = k - 1;
  }
  for (int x : answer) {
    cout << x << ' ';
  }
  return 0;
}

```