#include <bits/stdc++.h>
using namespace std;
const double PI = 3.14159265358979323;
const long long mod = 1000000000 + 7;
const long long INF = 1e18;
const long long P1 = 317;
const long long P2 = 239;
const long long N = 2e5;
const double kd = 1.41421356237;
vector<int> tree;
void Set(int v, int L, int R, int pos) {
  if (R - L == 1) {
    tree[v] = 1;
    return;
  }
  if (pos < (L + R) / 2)
    Set(v * 2, L, (L + R) / 2, pos);
  else
    Set(v * 2 + 1, (L + R) / 2, R, pos);
  tree[v] = tree[v * 2] + tree[v * 2 + 1];
}
pair<int, int> Get(int v, int L, int R, int k) {
  if (tree[v] < k) return {tree[v], -1};
  if (R - L == 1) return {1, L};
  pair<int, int> an1 = Get(v * 2, L, (L + R) / 2, k), an2;
  if (an1.first < k) {
    an2 = Get(v * 2 + 1, (L + R) / 2, R, k - an1.first);
    return {an2.first + an1.first, an2.second};
  }
  return an1;
}
int an[N];
pair<pair<int, int>, int> b[N];
int ff(pair<int, int> a, pair<int, int> b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second < b.second;
}
int ff2(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, mm;
  cin >> n;
  vector<pair<int, int> > a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  tree.resize(n * 4);
  cin >> mm;
  for (int i = 0; i < mm; i++) {
    cin >> b[i].first.first >> b[i].first.second;
    b[i].second = i;
  }
  sort(b, b + mm);
  sort(a.begin(), a.end(), ff);
  int j = 0, kk = 1;
  for (int i = 0; i < n; i++) {
    int x = a[i].second;
    Set(1, 0, n, x);
    while (b[j].first.first == kk) {
      int pos = b[j].first.second;
      int qq = b[j].second;
      j++;
      an[qq] = Get(1, 0, n, pos).second;
    }
    kk++;
  }
  sort(a.begin(), a.end(), ff2);
  for (int i = 0; i < mm; i++) cout << a[an[i]].first << '\n';
}

### 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;
const double PI = 3.14159265358979323;
const long long mod = 1000000000 + 7;
const long long INF = 1e18;
const long long P1 = 317;
const long long P2 = 239;
const long long N = 2e5;
const double kd = 1.41421356237;
vector<int> tree;
void Set(int v, int L, int R, int pos) {
  if (R - L == 1) {
    tree[v] = 1;
    return;
  }
  if (pos < (L + R) / 2)
    Set(v * 2, L, (L + R) / 2, pos);
  else
    Set(v * 2 + 1, (L + R) / 2, R, pos);
  tree[v] = tree[v * 2] + tree[v * 2 + 1];
}
pair<int, int> Get(int v, int L, int R, int k) {
  if (tree[v] < k) return {tree[v], -1};
  if (R - L == 1) return {1, L};
  pair<int, int> an1 = Get(v * 2, L, (L + R) / 2, k), an2;
  if (an1.first < k) {
    an2 = Get(v * 2 + 1, (L + R) / 2, R, k - an1.first);
    return {an2.first + an1.first, an2.second};
  }
  return an1;
}
int an[N];
pair<pair<int, int>, int> b[N];
int ff(pair<int, int> a, pair<int, int> b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second < b.second;
}
int ff2(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, mm;
  cin >> n;
  vector<pair<int, int> > a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  tree.resize(n * 4);
  cin >> mm;
  for (int i = 0; i < mm; i++) {
    cin >> b[i].first.first >> b[i].first.second;
    b[i].second = i;
  }
  sort(b, b + mm);
  sort(a.begin(), a.end(), ff);
  int j = 0, kk = 1;
  for (int i = 0; i < n; i++) {
    int x = a[i].second;
    Set(1, 0, n, x);
    while (b[j].first.first == kk) {
      int pos = b[j].first.second;
      int qq = b[j].second;
      j++;
      an[qq] = Get(1, 0, n, pos).second;
    }
    kk++;
  }
  sort(a.begin(), a.end(), ff2);
  for (int i = 0; i < mm; i++) cout << a[an[i]].first << '\n';
}

```