#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct BIT {
 private:
  vector<T> data;
  const int n;

 public:
  BIT(int sz) : data(sz + 1, 0), n(sz) {}
  BIT(vector<T> t) : data(t), n((int)t.size()) {}
  T sum(size_t i) {
    T s = 0;
    while (i > 0) {
      s += data[i];
      i -= i & (-i);
    }
    return s;
  }
  T sum(size_t i, size_t j) { return sum(j) - sum(i); }
  void add(int i, T x) {
    ++i;
    assert(i > 0);
    while (i <= n) {
      data[i] += x;
      i += i & (-i);
    }
  }
};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<long long> A(n);
  vector<pair<long long, int>> V;
  for (int i = 0; i < n; ++i) cin >> A[i], V.emplace_back(-A[i], i);
  sort(V.begin(), V.end());
  int m;
  cin >> m;
  vector<int> K(m), P(m);
  vector<vector<int>> I(n);
  for (int i = 0; i < m; ++i) {
    cin >> K[i] >> P[i];
    --K[i], --P[i];
    I[K[i]].push_back(i);
  }
  vector<int> ans(m, -1);
  BIT<int> bit(n + 1);
  for (int i = 0; i < n; ++i) {
    auto p = V[i];
    bit.add(p.second, 1);
    for (auto idx : I[i]) {
      int pos = P[idx];
      int ok = 0, ng = n;
      while (ng - ok > 1) {
        int mid = (ok + ng) / 2;
        if (bit.sum(mid) <= pos)
          ok = mid;
        else
          ng = mid;
      }
      ans[idx] = A[ok];
    }
  }
  for (auto a : ans) cout << a << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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]. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct BIT {
 private:
  vector<T> data;
  const int n;

 public:
  BIT(int sz) : data(sz + 1, 0), n(sz) {}
  BIT(vector<T> t) : data(t), n((int)t.size()) {}
  T sum(size_t i) {
    T s = 0;
    while (i > 0) {
      s += data[i];
      i -= i & (-i);
    }
    return s;
  }
  T sum(size_t i, size_t j) { return sum(j) - sum(i); }
  void add(int i, T x) {
    ++i;
    assert(i > 0);
    while (i <= n) {
      data[i] += x;
      i += i & (-i);
    }
  }
};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<long long> A(n);
  vector<pair<long long, int>> V;
  for (int i = 0; i < n; ++i) cin >> A[i], V.emplace_back(-A[i], i);
  sort(V.begin(), V.end());
  int m;
  cin >> m;
  vector<int> K(m), P(m);
  vector<vector<int>> I(n);
  for (int i = 0; i < m; ++i) {
    cin >> K[i] >> P[i];
    --K[i], --P[i];
    I[K[i]].push_back(i);
  }
  vector<int> ans(m, -1);
  BIT<int> bit(n + 1);
  for (int i = 0; i < n; ++i) {
    auto p = V[i];
    bit.add(p.second, 1);
    for (auto idx : I[i]) {
      int pos = P[idx];
      int ok = 0, ng = n;
      while (ng - ok > 1) {
        int mid = (ok + ng) / 2;
        if (bit.sum(mid) <= pos)
          ok = mid;
        else
          ng = mid;
      }
      ans[idx] = A[ok];
    }
  }
  for (auto a : ans) cout << a << endl;
}

```