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 N = 2e5 + 7;
const int eps = 1e-10;
const long long INF = 1e9 + 10;
long long t[4 * N];
long long a[N], sz = 1, n, q;
void build() {
  while (sz < n) {
    sz *= 2;
  }
}
void upd(int i, int x) {
  i += sz;
  t[i] = x;
  for (i /= 2; i > 0; i /= 2) {
    t[i] = t[i * 2] + t[i * 2 + 1];
  }
}
long long Sum(int l, int r) {
  long long i = l + sz, j = r + sz, sum = 0;
  while (i <= j) {
    if (i % 2) sum += t[i++];
    if (!(j % 2)) sum += t[j--];
    i /= 2;
    j /= 2;
  }
  return sum;
}
int m, l, r, x, k, pos;
string s;
vector<pair<int, int>> v;
set<int> st;
int ans[N];
void B_Search(int x, int i) {
  int l = -1, r = sz;
  while (l + 1 < r) {
    int m = (l + r) / 2;
    if (Sum(0, m) < x)
      l = m;
    else
      r = m;
  }
  ans[i] = a[r];
}
int main() {
  cin >> n;
  build();
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    st.insert(a[i]);
    v.push_back({a[i], i});
  }
  sort(v.begin(), v.end(), [](pair<int, int> a, pair<int, int> b) {
    if (a.first == b.first) return a.second < b.second;
    return a.first > b.first;
  });
  cin >> m;
  vector<pair<pair<int, int>, int>> q;
  for (int i = 0; i < m; ++i) {
    int k, pos;
    cin >> k >> pos;
    q.push_back({{k, pos}, i});
  }
  sort(q.begin(), q.end(),
       [](pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
         return a.first.first < b.first.first;
       });
  int r = 0;
  for (int i = 0; i < n; ++i) {
    upd(v[i].second, 1);
    while (r < m && q[r].first.first == i + 1) {
      B_Search(q[r].first.second, q[r].second);
      r++;
    }
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << endl;
  return 0;
}
