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 = 1e18;
struct query {
  long long k, pos, ind;
};
struct fenwick {
  long long n;
  vector<long long> f;
  fenwick(long long _n) {
    n = _n;
    f.resize(n, 0);
  }
  void add(long long val, long long i) {
    for (; i < n; i |= (i + 1)) f[i] += val;
  }
  long long sum(long long i) {
    long long ans = 0;
    for (; i >= 0; i &= (i + 1), --i) ans += f[i];
    return ans;
  }
};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  vector<long long> start(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i].first;
    start[i] = a[i].first;
    a[i].second = i;
  }
  sort(a.begin(), a.end(),
       [](pair<long long, long long> a, pair<long long, long long> b) {
         if (a.first == b.first) return a.second > b.second;
         return a.first < b.first;
       });
  reverse(a.begin(), a.end());
  long long q;
  cin >> q;
  vector<query> qs(q);
  for (long long t = 0; t < q; ++t) {
    cin >> qs[t].k >> qs[t].pos;
    qs[t].k--;
    qs[t].pos--;
    qs[t].ind = t;
  }
  sort(qs.begin(), qs.end(), [](query a, query b) { return a.k < b.k; });
  fenwick f(n);
  long long l = -1;
  vector<long long> ans(q);
  for (long long i = 0; i < q; ++i) {
    while (l < qs[i].k) {
      ++l;
      f.add(1, a[l].second);
    }
    long long L = 0, R = n - 1;
    while (R - L > 1) {
      long long m = (R + L) / 2;
      if (f.sum(m) <= qs[i].pos) {
        L = m;
      } else
        R = m;
    }
    if (f.sum(L) < qs[i].pos + 1) swap(L, R);
    ans[qs[i].ind] = start[L];
  }
  for (auto x : ans) cout << x << '\n';
}
