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;
vector<vector<int>> t;
int q(int v, int l, int r, int L, int R, int k) {
  if (l > R || r < L || L > R) return 0;
  if (l <= L && R <= r)
    return lower_bound(t[v].begin(), t[v].end(), k) - t[v].begin();
  return q(v * 2, l, r, L, (L + R) / 2, k) +
         q(v * 2 + 1, l, r, (L + R) / 2 + 1, R, k);
}
int32_t main() {
  int n, m, U = 1;
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  vector<pair<int, int>> a(n);
  vector<int> u(n);
  for (int i = 0; i < n; i++)
    cin >> a[i].first, a[i].second = -i, u[i] = a[i].first;
  while (U < n) U <<= 1;
  n = U;
  while (a.size() < n) a.push_back({-1e9, -(int)a.size()}), u.push_back(1e9);
  t.resize(n * 2);
  vector<int> z(n + 1);
  sort(a.rbegin(), a.rend());
  for (int i = 0; i < n; i++) z[i] = -a[i].second, t[i + n].push_back(z[i]);
  for (int i = n - 1; i > 0; i--) {
    for (auto e : t[i * 2]) t[i].push_back(e);
    for (auto e : t[i * 2 + 1]) t[i].push_back(e);
    sort(t[i].begin(), t[i].end());
  }
  cin >> m;
  while (m--) {
    int a, b, l = 0, r = n + 1;
    cin >> a >> b;
    while (l < r - 1) {
      int m = (l + r) / 2;
      int ans = q(1, 0, a - 1, 0, n - 1, m);
      if (ans > b - 1)
        r = m;
      else
        l = m;
    }
    cout << u[l] << "\n";
  }
}
