#include <bits/stdc++.h>
using namespace std;
long long int n;
vector<long long int> a, b;
struct segmentTree {
  vector<long long int> t;
  segmentTree() { t.resize(4 * n); }
  void add(long long int ind, long long int x, long long int v = 0,
           long long int tl = 0, long long int tr = n) {
    if (tr - tl == 1)
      t[v] += x;
    else {
      long long int mid = (tl + tr) / 2;
      if (ind >= mid)
        add(ind, x, 2 * v + 2, mid, tr);
      else
        add(ind, x, 2 * v + 1, tl, mid);
      t[v] = t[2 * v + 1] + t[2 * v + 2];
    }
  }
  long long int getK(long long int k, long long int v = 0, long long int tl = 0,
                     long long int tr = n) {
    if (tr - tl == 1)
      return tl;
    else {
      long long int mid = (tl + tr) / 2;
      if (t[2 * v + 1] >= k)
        return getK(k, 2 * v + 1, tl, mid);
      else
        return getK(k - t[2 * v + 1], 2 * v + 2, mid, tr);
    }
  }
};
struct query {
  long long int k, p, ind;
  query(long long int k = 0, long long int p = 0, long long int ind = 0)
      : k(k), p(p), ind(ind) {}
};
signed main() {
  cin >> n;
  a.resize(n);
  b.resize(n);
  iota(b.begin(), b.end(), 0);
  for (long long int i = 0; i < n; ++i) cin >> a[i];
  sort(b.begin(), b.end(), [](long long int i, long long int j) {
    return a[i] > a[j] || (a[i] == a[j] && i < j);
  });
  long long int q;
  cin >> q;
  vector<long long int> ans(q);
  vector<query> queries;
  for (long long int i = 0; i < q; ++i) {
    long long int k, p;
    cin >> k >> p;
    queries.emplace_back(k, p, i);
  }
  sort(queries.begin(), queries.end(),
       [](const query& a, const query& b) { return a.k < b.k; });
  long long int last = 0;
  segmentTree t;
  for (auto ev : queries) {
    while (last < ev.k) t.add(b[last++], 1);
    long long int ind = t.getK(ev.p);
    ans[ev.ind] = a[ind];
  }
  for (auto i : ans) cout << i << "\n";
  return 0;
}

### Prompt

Please create a solution in Cpp to 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 int n;
vector<long long int> a, b;
struct segmentTree {
  vector<long long int> t;
  segmentTree() { t.resize(4 * n); }
  void add(long long int ind, long long int x, long long int v = 0,
           long long int tl = 0, long long int tr = n) {
    if (tr - tl == 1)
      t[v] += x;
    else {
      long long int mid = (tl + tr) / 2;
      if (ind >= mid)
        add(ind, x, 2 * v + 2, mid, tr);
      else
        add(ind, x, 2 * v + 1, tl, mid);
      t[v] = t[2 * v + 1] + t[2 * v + 2];
    }
  }
  long long int getK(long long int k, long long int v = 0, long long int tl = 0,
                     long long int tr = n) {
    if (tr - tl == 1)
      return tl;
    else {
      long long int mid = (tl + tr) / 2;
      if (t[2 * v + 1] >= k)
        return getK(k, 2 * v + 1, tl, mid);
      else
        return getK(k - t[2 * v + 1], 2 * v + 2, mid, tr);
    }
  }
};
struct query {
  long long int k, p, ind;
  query(long long int k = 0, long long int p = 0, long long int ind = 0)
      : k(k), p(p), ind(ind) {}
};
signed main() {
  cin >> n;
  a.resize(n);
  b.resize(n);
  iota(b.begin(), b.end(), 0);
  for (long long int i = 0; i < n; ++i) cin >> a[i];
  sort(b.begin(), b.end(), [](long long int i, long long int j) {
    return a[i] > a[j] || (a[i] == a[j] && i < j);
  });
  long long int q;
  cin >> q;
  vector<long long int> ans(q);
  vector<query> queries;
  for (long long int i = 0; i < q; ++i) {
    long long int k, p;
    cin >> k >> p;
    queries.emplace_back(k, p, i);
  }
  sort(queries.begin(), queries.end(),
       [](const query& a, const query& b) { return a.k < b.k; });
  long long int last = 0;
  segmentTree t;
  for (auto ev : queries) {
    while (last < ev.k) t.add(b[last++], 1);
    long long int ind = t.getK(ev.p);
    ans[ev.ind] = a[ind];
  }
  for (auto i : ans) cout << i << "\n";
  return 0;
}

```