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;
ifstream in;
ofstream out;
const long long kk = 1000;
const long long ml = kk * kk;
const long long mod = ml * kk + 7;
const long long inf = ml * ml * ml + 7;
long long n, i, j;
vector<pair<long long, long long>> m;
vector<long long> p, rm;
vector<pair<pair<long long, long long>, long long>> qr;
long long done;
vector<long long> ans;
bool viv = false;
vector<long long> t;
long long s;
void show() {
  for (long long i = 1; i < 2 * s; i++) {
    cout << t[i] << ' ';
    if (!((i) & (i + 1))) cout << endl;
  }
  cout << "____ " << endl;
}
void build() {
  s = 1;
  while (s < n + 1) s <<= 1;
  t.resize(2 * s);
}
void prepare() {
  long long q;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long a, b;
    cin >> a >> b;
    qr.push_back({{a, b}, i});
  }
  sort(qr.begin(), qr.end());
  ans.resize(q, -1);
  build();
}
void add(long long pl) {
  pl += s;
  while (pl) {
    t[pl]++;
    pl >>= 1;
  }
}
long long find(long long v, long long l, long long r, long long k) {
  if (l == r) return r;
  long long mi = l + r >> 1;
  if (viv)
    cout << "I'm on " << v << " [" << l << "; " << r << "] and " << t[2 * v]
         << endl;
  if (t[2 * v] < k) return find(2 * v + 1, mi + 1, r, k - t[2 * v]);
  return find(2 * v, l, mi, k);
}
long long get(long long k) {
  if (viv) cout << "Hmmm " << find(1, 0, s - 1, k) << endl;
  return find(1, 0, s - 1, k);
}
void solve(pair<pair<long long, long long>, long long> z) {
  while (done < z.first.first) {
    add(p[done]);
    done++;
  }
  if (viv) {
    cout << "done " << done << endl;
    show();
  }
  long long res = get(z.first.second);
  if (viv) cout << "Here get the ans on " << res << endl;
  ans[z.second] = rm[res];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (i = 0; i < n; i++) {
    long long a;
    cin >> a;
    rm.push_back(a);
    m.push_back({a, -i});
  }
  sort(m.rbegin(), m.rend());
  for (auto i : m) p.push_back(-i.second);
  prepare();
  for (auto i : qr) solve(i);
  for (auto i : ans) cout << i << '\n';
  return 0;
}
