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;
int a[200002];
int ca[200003];
vector<pair<int, pair<int, int> > > input;
unordered_map<int, int> cmp, rcmp;
class OS {
 public:
  int segtree[800002];
  int query(int node, int l, int r, int ql, int qr) {
    if (qr < l || ql > r) return 0;
    if (ql <= l && qr >= r) return segtree[node];
    int m = (l + r) / 2;
    int a = query(2 * node, l, m, ql, qr);
    int b = query(2 * node + 1, m + 1, r, ql, qr);
    return a + b;
  }
  void update(int node, int l, int r, int in, int val) {
    if (l == r) {
      segtree[node] += val;
      return;
    }
    int m = (l + r) / 2;
    if (in <= m)
      update(2 * node, l, m, in, val);
    else
      update(2 * node + 1, m + 1, r, in, val);
    segtree[node] = segtree[2 * node] + segtree[2 * node + 1];
  }
  int ranking(int x, int n) { return query(1, 1, n, 1, x - 1); }
  int order_stats(int k, int n) {
    int l = 1;
    int r = n + 1;
    while (l < r) {
      int m = (l + r) / 2;
      if (ranking(m, n) < k)
        l = m + 1;
      else
        r = m;
    }
    return l - 1;
  }
};
int ans[200003];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  set<int> s;
  OS os = OS();
  for (int i = 1; i <= n; i++) cin >> a[i], s.insert(a[i]), rcmp[i] = a[i];
  int in = 1;
  for (auto x : s) {
    cmp[x] = in;
    in++;
  }
  for (int i = 1; i <= n; i++) a[i] = cmp[a[i]];
  int m;
  cin >> m;
  int tm = m;
  int mind = 1;
  while (tm--) {
    int i, j;
    cin >> i >> j;
    input.push_back({i, {j, mind}});
    mind++;
  }
  sort(input.begin(), input.end());
  vector<pair<int, int> > v;
  for (int i = 1; i <= n; i++) {
    v.push_back({-a[i], i});
  }
  sort(v.begin(), v.end());
  int iter_k = 0;
  for (auto p : input) {
    int k = p.first;
    int qin = p.second.first;
    int inputin = p.second.second;
    while (iter_k + 1 <= k) {
      auto pp = v[iter_k];
      int val = -pp.first;
      int in = pp.second;
      os.update(1, 1, n, in, 1);
      iter_k++;
    }
    int pans = os.order_stats(qin, n);
    ans[inputin] = pans;
  }
  for (int i = 1; i <= m; i++) cout << rcmp[ans[i]] << "\n";
  return 0;
}
