#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
template <int N, typename T = int>
struct fenwick_multiset {
  T s = 0, tree[N + 1];
  T size() const { return s; }
  fenwick_multiset() { fill(tree, tree + N + 1, 0); }
  void insert(int x, const T k = 1) {
    for (x++; x <= N; x += x & -x) tree[x] += k;
    s += k;
  }
  void remove(int x, const T k = 1) {
    for (x++; x <= N; x += x & -x) tree[x] -= k;
    s -= k;
  }
  T index(int x) const {
    if (x <= 0) return 0;
    if (x >= N) return size();
    T i = 0;
    for (; x > 0; x -= x & -x) i += tree[x];
    return i;
  }
  T count(int x) const {
    if (x < 0 || x >= N) return 0;
    int y = x + 1;
    T i = tree[x + 1];
    for (y -= y & -y; x != y; x -= x & -x) i -= tree[x];
    return i;
  }
  int get(T x) const {
    int i = 0;
    for (int m = 1 << (31 - __builtin_clz(N)); m > 0; m >>= 1)
      if (i + m <= N && tree[i + m] <= x) x -= tree[i += m];
    return i;
  }
  int next_greater(int x) { return get(index(x + 1)); }
  int next_lower(int x) {
    T i = index(x);
    return i == 0 ? -1 : get(i - 1);
  }
  T count_interval(int l, int r) { return index(r) - index(l); }
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int n, m;
  cin >> n;
  set<int> nums;
  vector<int> nn(n);
  vector<pair<int, int>> ord;
  vector<tuple<int, int, int>> qr;
  for (int i = 0, v; i < n; ++i) {
    cin >> v;
    nn[i] = v;
    nums.insert(v);
    ord.push_back({v, -i});
  }
  cin >> m;
  vector<int> ans(m);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    --v;
    qr.push_back({u, v, i});
  }
  sort((qr).begin(), (qr).end());
  sort((ord).rbegin(), (ord).rend());
  map<int, int> mp;
  map<int, int> rv;
  int idx;
  for (auto i : nums) {
    mp[i] = idx;
    rv[idx] = i;
    ++idx;
  }
  for (int i = 0; i < n; ++i) {
    ord[i].second *= -1;
  }
  fenwick_multiset<200001, int> ff;
  int ss = 0;
  for (auto [k, id, i] : qr) {
    while (ss != k) ff.insert(ord[ss++].second);
    ans[i] = nn[ff.get(id)];
  }
  for (auto i : ans) {
    cout << i << 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;
#pragma GCC optimize("O3")
template <int N, typename T = int>
struct fenwick_multiset {
  T s = 0, tree[N + 1];
  T size() const { return s; }
  fenwick_multiset() { fill(tree, tree + N + 1, 0); }
  void insert(int x, const T k = 1) {
    for (x++; x <= N; x += x & -x) tree[x] += k;
    s += k;
  }
  void remove(int x, const T k = 1) {
    for (x++; x <= N; x += x & -x) tree[x] -= k;
    s -= k;
  }
  T index(int x) const {
    if (x <= 0) return 0;
    if (x >= N) return size();
    T i = 0;
    for (; x > 0; x -= x & -x) i += tree[x];
    return i;
  }
  T count(int x) const {
    if (x < 0 || x >= N) return 0;
    int y = x + 1;
    T i = tree[x + 1];
    for (y -= y & -y; x != y; x -= x & -x) i -= tree[x];
    return i;
  }
  int get(T x) const {
    int i = 0;
    for (int m = 1 << (31 - __builtin_clz(N)); m > 0; m >>= 1)
      if (i + m <= N && tree[i + m] <= x) x -= tree[i += m];
    return i;
  }
  int next_greater(int x) { return get(index(x + 1)); }
  int next_lower(int x) {
    T i = index(x);
    return i == 0 ? -1 : get(i - 1);
  }
  T count_interval(int l, int r) { return index(r) - index(l); }
};
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int n, m;
  cin >> n;
  set<int> nums;
  vector<int> nn(n);
  vector<pair<int, int>> ord;
  vector<tuple<int, int, int>> qr;
  for (int i = 0, v; i < n; ++i) {
    cin >> v;
    nn[i] = v;
    nums.insert(v);
    ord.push_back({v, -i});
  }
  cin >> m;
  vector<int> ans(m);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    --v;
    qr.push_back({u, v, i});
  }
  sort((qr).begin(), (qr).end());
  sort((ord).rbegin(), (ord).rend());
  map<int, int> mp;
  map<int, int> rv;
  int idx;
  for (auto i : nums) {
    mp[i] = idx;
    rv[idx] = i;
    ++idx;
  }
  for (int i = 0; i < n; ++i) {
    ord[i].second *= -1;
  }
  fenwick_multiset<200001, int> ff;
  int ss = 0;
  for (auto [k, id, i] : qr) {
    while (ss != k) ff.insert(ord[ss++].second);
    ans[i] = nn[ff.get(id)];
  }
  for (auto i : ans) {
    cout << i << endl;
  }
  return 0;
}

```