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 long long inf = 1e9 + 7;
const long long INF = 1LL << 60;
const long long mod = 1e9 + 7;
const long double eps = 1e-8;
const long double pi = acos(-1.0);
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class Abel>
struct BIT {
 private:
  vector<Abel> node;
  long long n;
  Abel UNITY_SUM = 0;

 public:
  BIT(long long n_) {
    n = n_;
    node.resize(n, UNITY_SUM);
  }
  void add(long long a, Abel w) {
    for (long long i = a; i < n; i |= i + 1) node[i] += w;
  }
  Abel sum(long long a) {
    Abel ret = UNITY_SUM;
    for (long long i = a - 1; i >= 0; i = (i & (i + 1)) - 1) ret += node[i];
    return ret;
  }
  Abel sum(long long a, long long b) { return sum(b) - sum(a); }
  long long get(long long k) {
    ++k;
    long long res = 0;
    long long n = 1;
    while (n < (long long)node.size()) n *= 2;
    for (long long i = n / 2; i > 0; i /= 2) {
      if (res + i <= (long long)node.size() && node[res + i - 1] < k) {
        k -= node[res + i - 1];
        res += i;
      }
    }
    return res;
  }
  void print() {
    for (long long i = 0; i < n; ++i) cout << sum(i, i + 1) << ",";
    cout << endl;
  }
};
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long m;
  cin >> m;
  vector<long long> ans(m);
  vector<vector<pair<long long, long long>>> g(
      n + 1, vector<pair<long long, long long>>());
  for (long long i = 0; i < m; i++) {
    long long k, pos;
    cin >> k >> pos;
    --pos;
    g[k].push_back({i, pos});
  }
  vector<pair<long long, long long>> x(n);
  for (long long i = 0; i < n; i++) x[i] = {a[i], -i};
  sort(x.begin(), x.end());
  reverse(x.begin(), x.end());
  BIT<long long> ids(n);
  for (long long k = 0; k < n; k++) {
    ids.add(-x[k].second, 1);
    for (auto p : g[k + 1]) {
      long long j = p.first, pos = p.second;
      ans[j] = a[ids.get(pos)];
    }
  }
  for (long long i = 0; i < m; i++) {
    cout << ans[i] << endl;
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
