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 = 200005;
struct Query {
  int k, pos, id;
} q[N];
vector<pair<int, int> > v;
int n, m, a[N], res[N], bit[N];
bool cmp(pair<int, int> first, pair<int, int> second) {
  if (first.first == second.first) return first.second < second.second;
  return first.first > second.first;
}
bool cmp2(Query first, Query second) { return first.k < second.k; }
struct OrderSet {
  struct Trie {
    int cnt;
    int next[2];
    Trie() {
      memset(next, -1, sizeof(next));
      cnt = 0;
    }
  };
  vector<Trie> trie;
  OrderSet() {
    auto u = Trie();
    trie.push_back(u);
  }
  bool find(int first) {
    int cur = 0;
    for (int i = 20; i >= 0; i--) {
      auto u = (first >> i) & 1;
      if (trie[cur].next[u] == -1)
        return 0;
      else
        cur = trie[cur].next[u];
    }
    return 1;
  }
  void insert(int first) {
    int cur = 0;
    trie[cur].cnt++;
    for (int i = 20; i >= 0; i--) {
      auto u = (first >> i) & 1;
      if (trie[cur].next[u] == -1) {
        auto newTrie = Trie();
        trie.push_back(newTrie);
        trie[cur].next[u] = trie.size() - 1;
        cur = trie[cur].next[u];
      } else {
        cur = trie[cur].next[u];
      }
      trie[cur].cnt++;
    }
  }
  void deletee(int first) {
    int cur = 0;
    trie[cur].cnt--;
    for (int i = 20; i >= 0; i--) {
      auto u = (first >> i) & 1;
      int temp = trie[cur].next[u];
      if (trie[temp].cnt == 1) trie[cur].next[u] = -1;
      cur = temp;
      trie[cur].cnt--;
    }
  }
  int order_of_key(int first) {
    int cur = 0;
    int res = 0;
    for (int i = 20; i >= 0; i--) {
      auto u = (first >> i) & 1;
      if (u == 1 && trie[cur].next[0] != -1) {
        int temp = trie[cur].next[0];
        res += trie[temp].cnt;
      }
      if (trie[cur].next[u] == -1) return res;
      cur = trie[cur].next[u];
    }
    return res;
  }
  int find_by_order(int k) {
    int cur = 0;
    int res = 0;
    if (trie[cur].cnt < k) return -1;
    for (int i = 20; i >= 0; i--) {
      int temp = trie[cur].next[0];
      if (temp == -1) {
        cur = trie[cur].next[1];
        res += (1 << i);
        continue;
      }
      if (trie[temp].cnt < k) {
        k -= trie[temp].cnt;
        cur = trie[cur].next[1];
        res += (1 << i);
      } else {
        cur = trie[cur].next[0];
      }
    }
    return res;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  OrderSet *os = new OrderSet();
  cin >> n;
  v.push_back({1e9 + 2, 1e9});
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    v.push_back({a[i], i});
  }
  sort(v.begin(), v.end(), cmp);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> q[i].k >> q[i].pos;
    q[i].id = i;
  }
  sort(q + 1, q + m + 1, cmp2);
  int cur = 0;
  for (int i = 1; i <= m; i++) {
    while (cur < q[i].k) {
      cur++;
      os->insert(v[cur].second);
    }
    res[q[i].id] = os->find_by_order(q[i].pos);
  }
  for (int i = 1; i <= m; i++) cout << a[res[i]] << '\n';
  return 0;
}
