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]. 
#include <bits/stdc++.h>
using namespace std;
const int N = 500023;
long long int vis[N] = {0};
vector<long long int> adj[N];
long long int count1[N] = {0};
bool sortbysec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  if (a.first > b.first)
    return 1;
  else if (a.first == b.first)
    return (a.second < b.second);
  else
    return 0;
}
long long int query(long long int l, long long int r, long long int pos,
                    long long int k) {
  if (count1[pos] < k) return -1;
  if (l == r) return l;
  long long int mid = (l + r) / 2;
  if (count1[2 * pos + 1] >= k)
    return query(l, mid, 2 * pos + 1, k);
  else
    return query(mid + 1, r, 2 * pos + 2, k - count1[2 * pos + 1]);
}
void update(long long int l, long long int r, long long int val,
            long long int pos) {
  if (val < l || val > r) return;
  if (l == r) {
    count1[pos] = 1;
    return;
  }
  long long int mid = (l + r) / 2;
  if (val <= mid)
    update(l, mid, val, 2 * pos + 1);
  else
    update(mid + 1, r, val, 2 * pos + 2);
  count1[pos] = count1[2 * pos + 1] + count1[2 * pos + 2];
}
void solve() {
  long long int n;
  cin >> n;
  vector<pair<long long int, long long int>> v;
  vector<long long int> copy;
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    v.push_back({x, i + 1});
    copy.push_back(x);
  }
  sort((v).begin(), (v).end(), sortbysec);
  long long int m;
  cin >> m;
  vector<pair<long long int, long long int>> q;
  map<pair<long long int, long long int>, long long int> ans;
  vector<pair<long long int, long long int>> order;
  for (long long int i = 0; i < m; i++) {
    long long int k, pos;
    cin >> k >> pos;
    q.push_back({k, pos});
    order.push_back({k, pos});
  }
  sort((q).begin(), (q).end());
  long long int curr = 0;
  for (long long int i = 0; i < m; i++) {
    for (long long int j = curr; j < q[i].first; j++)
      update(0, n + 1, v[j].second, 0);
    curr = q[i].first;
    ans[{q[i].first, q[i].second}] = query(0, n + 1, 0, q[i].second);
  }
  for (long long int i = 0; i < m; i++) cout << copy[ans[order[i]] - 1] << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int T;
  T = 1;
  long long int t = 0;
  while (t++ < T) {
    solve();
    cout << '\n';
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}
