#include <bits/stdc++.h>
using namespace std;
int n, h, k, i, x, y, t[800000];
vector<int> v1;
vector<pair<int, int>> v, v2;
vector<pair<pair<int, int>, int>> q;
bool comp(pair<int, int> a, pair<int, int> b) {
  return a.second > b.second || a.second == b.second && a.first < b.first;
}
bool comp1(pair<int, int> a, pair<int, int> b) { return a.first < b.first; }
void build(int p, int l, int r) {
  if (l + 1 == r) {
    t[p] = 0;
    return;
  }
  int m = (l + r) / 2;
  build(2 * p + 1, l, m);
  build(2 * p + 2, m, r);
  t[p] = t[2 * p + 1] + t[2 * p + 2];
}
void update(int p, int l, int r, int index, int v = 1) {
  if (index < l || r <= index) return;
  if (l + 1 == r) {
    t[p] = v;
    return;
  }
  int m = (l + r) / 2;
  update(2 * p + 1, l, m, index, v);
  update(2 * p + 2, m, r, index, v);
  t[p] = t[2 * p + 1] + t[2 * p + 2];
}
void get(int l, int r, int index, int v) {
  int p = 0, s = 0;
  while (l + 1 < r) {
    if (t[2 * p + 1] + s >= v) {
      p = 2 * p + 1;
      r = (l + r) / 2;
    } else {
      s += t[2 * p + 1];
      p = 2 * p + 2;
      l = (l + r) / 2;
    }
  }
  v2.push_back({index, l});
}
int main() {
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    v.push_back({i, x});
    v1.push_back(x);
  }
  sort(v.begin(), v.end(), comp);
  cin >> h;
  for (int j = 0; j < h; j++) {
    cin >> x >> y;
    q.push_back({{x, y}, j});
  }
  sort(q.begin(), q.end());
  build(0, 0, n);
  k = 0;
  for (int j = 0; j < h; j++) {
    while (k < q[j].first.first) {
      update(0, 0, n, v[k].first);
      k++;
    }
    get(0, n, q[j].second, q[j].first.second);
  }
  sort(v2.begin(), v2.end(), comp1);
  for (int j = 0; j < h; j++) cout << v1[v2[j].second] << endl;
  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;
int n, h, k, i, x, y, t[800000];
vector<int> v1;
vector<pair<int, int>> v, v2;
vector<pair<pair<int, int>, int>> q;
bool comp(pair<int, int> a, pair<int, int> b) {
  return a.second > b.second || a.second == b.second && a.first < b.first;
}
bool comp1(pair<int, int> a, pair<int, int> b) { return a.first < b.first; }
void build(int p, int l, int r) {
  if (l + 1 == r) {
    t[p] = 0;
    return;
  }
  int m = (l + r) / 2;
  build(2 * p + 1, l, m);
  build(2 * p + 2, m, r);
  t[p] = t[2 * p + 1] + t[2 * p + 2];
}
void update(int p, int l, int r, int index, int v = 1) {
  if (index < l || r <= index) return;
  if (l + 1 == r) {
    t[p] = v;
    return;
  }
  int m = (l + r) / 2;
  update(2 * p + 1, l, m, index, v);
  update(2 * p + 2, m, r, index, v);
  t[p] = t[2 * p + 1] + t[2 * p + 2];
}
void get(int l, int r, int index, int v) {
  int p = 0, s = 0;
  while (l + 1 < r) {
    if (t[2 * p + 1] + s >= v) {
      p = 2 * p + 1;
      r = (l + r) / 2;
    } else {
      s += t[2 * p + 1];
      p = 2 * p + 2;
      l = (l + r) / 2;
    }
  }
  v2.push_back({index, l});
}
int main() {
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    v.push_back({i, x});
    v1.push_back(x);
  }
  sort(v.begin(), v.end(), comp);
  cin >> h;
  for (int j = 0; j < h; j++) {
    cin >> x >> y;
    q.push_back({{x, y}, j});
  }
  sort(q.begin(), q.end());
  build(0, 0, n);
  k = 0;
  for (int j = 0; j < h; j++) {
    while (k < q[j].first.first) {
      update(0, 0, n, v[k].first);
      k++;
    }
    get(0, n, q[j].second, q[j].first.second);
  }
  sort(v2.begin(), v2.end(), comp1);
  for (int j = 0; j < h; j++) cout << v1[v2[j].second] << endl;
  return 0;
}

```