#include <bits/stdc++.h>
using namespace std;
mt19937 rnd;
struct node {
  int x = 0;
  int y = 0;
  int sz = 1;
  node *left = nullptr;
  node *right = nullptr;
  node() { y = rnd(); }
  node(int val) {
    y = rnd();
    x = val;
  }
};
int sz(node *r) {
  if (!r) {
    return 0;
  }
  return r->sz;
}
void relax(node *r) {
  if (!r) {
    return;
  }
  r->sz = sz(r->left) + sz(r->right) + 1;
}
node *merge(node *r1, node *r2) {
  if (!r1) {
    return r2;
  }
  if (!r2) {
    return r1;
  }
  if (r1->y > r2->y) {
    r1->right = merge(r1->right, r2);
    relax(r1);
    return r1;
  } else {
    r2->left = merge(r1, r2->left);
    relax(r2);
    return r2;
  }
}
pair<node *, node *> split(node *r, int x) {
  if (!r) {
    return {nullptr, nullptr};
  }
  if (r->x < x) {
    auto sup = split(r->right, x);
    r->right = sup.first;
    relax(r);
    return {r, sup.second};
  } else {
    auto sup = split(r->left, x);
    r->left = sup.second;
    relax(r);
    return {sup.first, r};
  }
}
int getkth(node *r, int k) {
  if (sz(r->left) == k) {
    return r->x;
  }
  if (sz(r->left) > k) {
    return getkth(r->left, k);
  } else {
    return getkth(r->right, k - sz(r->left) - 1);
  }
}
node *add(node *r, int val) {
  auto sup = split(r, val);
  node *n = new node(val);
  return merge(merge(sup.first, n), sup.second);
}
bool cmp(pair<int, int> a, pair<int, int> b) {
  return (a.first > b.first || (a.first == b.first && a.second < b.second));
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<pair<int, int> > c;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    c.push_back({a[i], i});
  }
  int m;
  cin >> m;
  node *r = nullptr;
  sort(c.begin(), c.end(), cmp);
  vector<int> ans(m);
  vector<vector<int> > q(m, vector<int>(3));
  for (int i = 0; i < m; i++) {
    cin >> q[i][0] >> q[i][1];
    q[i][2] = i;
    q[i][1]--;
  }
  sort(q.begin(), q.end());
  int now = 0;
  for (int i = 0; i < m; i++) {
    while (now < q[i][0]) {
      r = add(r, c[now].second);
      now++;
    }
    ans[q[i][2]] = a[getkth(r, q[i][1])];
  }
  for (auto i : ans) {
    cout << i << "\n";
  }
  cout << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
mt19937 rnd;
struct node {
  int x = 0;
  int y = 0;
  int sz = 1;
  node *left = nullptr;
  node *right = nullptr;
  node() { y = rnd(); }
  node(int val) {
    y = rnd();
    x = val;
  }
};
int sz(node *r) {
  if (!r) {
    return 0;
  }
  return r->sz;
}
void relax(node *r) {
  if (!r) {
    return;
  }
  r->sz = sz(r->left) + sz(r->right) + 1;
}
node *merge(node *r1, node *r2) {
  if (!r1) {
    return r2;
  }
  if (!r2) {
    return r1;
  }
  if (r1->y > r2->y) {
    r1->right = merge(r1->right, r2);
    relax(r1);
    return r1;
  } else {
    r2->left = merge(r1, r2->left);
    relax(r2);
    return r2;
  }
}
pair<node *, node *> split(node *r, int x) {
  if (!r) {
    return {nullptr, nullptr};
  }
  if (r->x < x) {
    auto sup = split(r->right, x);
    r->right = sup.first;
    relax(r);
    return {r, sup.second};
  } else {
    auto sup = split(r->left, x);
    r->left = sup.second;
    relax(r);
    return {sup.first, r};
  }
}
int getkth(node *r, int k) {
  if (sz(r->left) == k) {
    return r->x;
  }
  if (sz(r->left) > k) {
    return getkth(r->left, k);
  } else {
    return getkth(r->right, k - sz(r->left) - 1);
  }
}
node *add(node *r, int val) {
  auto sup = split(r, val);
  node *n = new node(val);
  return merge(merge(sup.first, n), sup.second);
}
bool cmp(pair<int, int> a, pair<int, int> b) {
  return (a.first > b.first || (a.first == b.first && a.second < b.second));
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<pair<int, int> > c;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    c.push_back({a[i], i});
  }
  int m;
  cin >> m;
  node *r = nullptr;
  sort(c.begin(), c.end(), cmp);
  vector<int> ans(m);
  vector<vector<int> > q(m, vector<int>(3));
  for (int i = 0; i < m; i++) {
    cin >> q[i][0] >> q[i][1];
    q[i][2] = i;
    q[i][1]--;
  }
  sort(q.begin(), q.end());
  int now = 0;
  for (int i = 0; i < m; i++) {
    while (now < q[i][0]) {
      r = add(r, c[now].second);
      now++;
    }
    ans[q[i][2]] = a[getkth(r, q[i][1])];
  }
  for (auto i : ans) {
    cout << i << "\n";
  }
  cout << endl;
  return 0;
}

```