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 MAXN = 2e5 + 10;
pair<int, int> t[4 * MAXN];
void update(int x, int vl, int vr, int pos, int val) {
  if (pos < vl || pos > vr)
    return;
  else if (vl == vr) {
    t[x].first++;
    t[x].second = val;
  } else {
    int m = (vl + vr) / 2;
    update(x * 2, vl, m, pos, val);
    update(x * 2 + 1, m + 1, vr, pos, val);
    t[x].first++;
  }
}
int srch(int x, int vl, int vr, int pos) {
  if (vl == vr) {
    return t[x].second;
  } else {
    int m = (vl + vr) / 2;
    int s1, s2;
    s1 = t[x * 2].first;
    s2 = t[x * 2 + 1].first;
    if (s1 >= pos) {
      srch(x * 2, vl, m, pos);
    } else {
      pos -= s1;
      srch(x * 2 + 1, m + 1, vr, pos);
    }
  }
}
void debug(int v, int vl, int vr) {
  cout << "v=" << v << "  vl=" << vl << "   vr=" << vr << "  t=" << t[v].first
       << ", " << t[v].second << endl;
  if (vl != vr) {
    int m = (vl + vr) / 2;
    debug(v * 2, vl, m);
    debug(v * 2 + 1, m + 1, vr);
  }
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int> > v(n);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    pair<int, int> p = {x, -i};
    v[i] = p;
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
  }
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int> > ask(m);
  vector<int> ans(m);
  int u = 0;
  while (m--) {
    int k, pos;
    cin >> k >> pos;
    pair<int, int> p = {k, pos};
    pair<pair<int, int>, int> q = {p, u};
    ask[u] = q;
    u++;
  }
  sort(ask.begin(), ask.end());
  int j = 0;
  for (int i = 1; i <= n; i++) {
    int ps = -v[i - 1].second;
    update(1, 0, MAXN, ps, v[i - 1].first);
    while (j < u && ask[j].first.first == i) {
      pair<int, int> tmp = ask[j].first;
      int d = ask[j].second;
      ans[d] = srch(1, 0, MAXN, tmp.second);
      j++;
    }
  }
  for (int i = 0; i < u; i++) cout << ans[i] << "\n";
}
int main() { solve(); }
