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;
template <typename A>
using VE = vector<A>;
template <typename A>
using USET = unordered_set<A>;
template <typename A>
using HEAP = priority_queue<A>;
template <typename A, typename B>
using PA = pair<A, B>;
template <typename A, typename B>
using UMAP = unordered_map<A, B>;
template <typename A>
using RHEAP = priority_queue<A, vector<A>, greater<A> >;
template <typename A>
A MAX(const A &a) {
  return a;
}
template <typename A>
A MIN(const A &a) {
  return a;
}
template <typename A>
A MAX(const A *a, const A *b) {
  return *max_element(a, b);
}
template <typename A>
A MIN(const A *a, const A *b) {
  return *min_element(a, b);
}
template <typename A, typename... B>
A MAX(const A &a, const B &...b) {
  return max(a, MAX(b...));
}
template <typename A, typename... B>
A MIN(const A &a, const B &...b) {
  return min(a, MIN(b...));
}
template <typename A, typename B = typename std::iterator_traits<A>::value_type>
B MAX(A a, A b) {
  return *max_element(a, b);
}
template <typename A, typename B = typename std::iterator_traits<A>::value_type>
B MIN(A a, A b) {
  return *min_element(a, b);
}
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
const int N = 200010, M = 1010;
int n, m;
int a[N];
int k[N], pos[N];
int order[N];
int tr[N];
int res[N];
vector<int> len[N];
void add(int x, int c) {
  for (int i = x; i < N; i += (i & -i)) tr[i] += c;
}
int find_k_min(int k) {
  int cnt = 0, x = 0;
  int MAX_VAL = N;
  for (int i = log2(MAX_VAL); i >= 0; i--) {
    x += (1 << i);
    if (x >= MAX_VAL || cnt + tr[x] >= k)
      x -= (1 << i);
    else
      cnt += tr[x];
  }
  return x + 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  iota(order + 1, order + 1 + n, 1);
  sort(order + 1, order + 1 + n, [&](int x, int y) {
    if (a[x] != a[y])
      return a[x] > a[y];
    else
      return x < y;
  });
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> k[i] >> pos[i];
    len[k[i]].emplace_back(i);
  }
  for (int i = 1; i <= n; i++) {
    add(order[i], 1);
    for (auto &j : len[i]) {
      int idx = find_k_min(pos[j]);
      res[j] = a[idx];
    }
  }
  for (int i = 1; i <= m; i++) cout << res[i] << '\n';
  return 0;
}
