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;
long long N;
long long n, x, k, z, m, pos, q, ans_v, y;
long long x1, p1, x2;
pair<long long, long long> a[200500];
vector<pair<long long, long long> > f;
struct SegmentTree {
  SegmentTree(long long n) : n(n), t(n * 2, 0) {}
  SegmentTree(vector<long long> x) : n(x.size()), t(n * 2, 0) {
    for (int i = 0; i < n; ++i) {
      t[i + n] = x[i];
    }
    build();
  }
  long long combine(long long a, long long b) { return a + b; }
  void build() {
    for (long long i = n - 1; i > 0; --i)
      t[i] = combine(t[i << 1], t[(i << 1) | 1]);
  }
  void modify(long long p, long long value) {
    for (t[p += n] = value; p > 1; p >>= 1) t[p >> 1] = combine(t[p], t[p ^ 1]);
  }
  long long query(long long l, long long r) {
    long long res = 0;
    for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
      if (l & 1) res = combine(res, t[l++]);
      if (r & 1) res = combine(res, t[--r]);
    }
    return res;
  }
  long long n;
  vector<long long> t;
};
long long ans[200500];
vector<pair<long long, long long> > queries[200500];
SegmentTree seg_tree(200500);
long long b[200500];
bool cmp(pair<long long, long long> x, pair<long long, long long> y) {
  return make_pair(-x.first, x.second) < make_pair(-y.first, y.second);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n, cmp);
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    queries[x].push_back(make_pair(y, i));
  }
  for (int i = 1; i <= n; i++) {
    b[a[i - 1].second] = a[i - 1].first;
    seg_tree.modify(a[i - 1].second, 1);
    for (auto curq : queries[i]) {
      long long l = 0, r = n - 1;
      long long val = curq.first;
      while (l < r) {
        long long mid = (l + r) / 2;
        if (seg_tree.query(0, mid + 1) >= val) {
          r = mid;
        } else {
          l = mid + 1;
        }
      }
      ans[curq.second] = b[l];
    }
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}
