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>
const long long oo = (long long)1e18 + 1;
using namespace std;
int n, q;
int b[200002];
pair<int, int> a[200002];
int T[200002];
vector<pair<int, int> > query[200002];
int k[200002], p[200002];
int res[200002];
void upd(int x) {
  while (x < 200002) {
    T[x]++;
    x = x + (x & (x ^ (x - 1)));
  }
}
int get(int x) {
  int res = 0;
  while (x > 0) {
    res += T[x];
    x = x - (x & (x ^ (x - 1)));
  }
  return res;
}
int find_kth(int cnt) {
  int l = 1;
  int r = n;
  int res = n;
  while (l <= r) {
    int mid = (l + r) >> 1;
    int count_value = get(mid);
    if (count_value >= cnt) {
      res = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  return res;
}
bool cmp(pair<int, int> x, pair<int, int> y) {
  return x.first > y.first || (x.first == y.first && x.second < y.second);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    b[i] = x;
    a[i] = make_pair(x, i);
  }
  sort(a + 1, a + n + 1, cmp);
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> k[i] >> p[i];
    query[k[i]].push_back(make_pair(p[i], i));
  }
  for (int i = 1; i <= n; i++) {
    upd(a[i].second);
    for (int j = 0; j < query[i].size(); j++) {
      pair<int, int> cur_query = query[i][j];
      int pos = cur_query.first;
      int rpos = cur_query.second;
      res[rpos] = b[find_kth(pos)];
    }
  }
  for (int i = 1; i <= q; i++) cout << res[i] << "\n";
  return 0;
}
