#include <bits/stdc++.h>
using namespace std;
ifstream in("input.in");
ofstream out("output.out");
const long long nmx = 2e5 + 5;
const long long inf = 1 << 30;
const long long MOD = 1e9 + 7;
const long long dx[4] = {-1, 0, 1, 0};
const long long dy[4] = {0, 1, 0, -1};
const string YES = "YES";
const string NO = "NO";
struct Query {
  long long k, p, index;
};
long long bit[nmx];
long long n, q, answer[nmx], arr[nmx];
vector<pair<long long, long long>> vec, aux;
vector<Query> query;
void add(long long val) {
  for (long long k = val; k <= 2e5; k += (k & -k)) bit[k] += 1;
}
long long sum(long long pos) {
  long long s = 0;
  for (long long k = pos; k > 0; k -= (k & -k)) s += bit[k];
  return s;
}
long long very_quick_select(long long kth) {
  long long l = 0;
  long long r = 2e5;
  while (l < r) {
    long long m = (l + r) >> 1;
    if (kth <= sum(m))
      r = m;
    else
      l = m + 1;
  }
  return l;
}
bool _comp(const pair<long long, long long> &A,
           const pair<long long, long long> &B) {
  return A.first < B.first || (A.first == B.first && A.second > B.second);
}
bool __comp(const Query &A, const Query &B) { return A.k < B.k; }
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    long long x;
    cin >> x;
    arr[i] = x;
    vec.push_back({x, i});
  }
  sort(vec.begin(), vec.end(), _comp);
  aux.push_back(vec.back());
  add(aux.back().second);
  vec.pop_back();
  cin >> q;
  for (long long i = 0; i < q; ++i) {
    long long k, p;
    cin >> k >> p;
    query.push_back({k, p, i});
  }
  sort(query.begin(), query.end(), __comp);
  for (long long i = 0; i < query.size(); ++i) {
    while (query[i].k > aux.size()) {
      aux.push_back(vec.back());
      add(aux.back().second);
      vec.pop_back();
    }
    answer[query[i].index] = arr[very_quick_select(query[i].p)];
  }
  for (long long i = 0; i < q; ++i) {
    cout << answer[i] << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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;
ifstream in("input.in");
ofstream out("output.out");
const long long nmx = 2e5 + 5;
const long long inf = 1 << 30;
const long long MOD = 1e9 + 7;
const long long dx[4] = {-1, 0, 1, 0};
const long long dy[4] = {0, 1, 0, -1};
const string YES = "YES";
const string NO = "NO";
struct Query {
  long long k, p, index;
};
long long bit[nmx];
long long n, q, answer[nmx], arr[nmx];
vector<pair<long long, long long>> vec, aux;
vector<Query> query;
void add(long long val) {
  for (long long k = val; k <= 2e5; k += (k & -k)) bit[k] += 1;
}
long long sum(long long pos) {
  long long s = 0;
  for (long long k = pos; k > 0; k -= (k & -k)) s += bit[k];
  return s;
}
long long very_quick_select(long long kth) {
  long long l = 0;
  long long r = 2e5;
  while (l < r) {
    long long m = (l + r) >> 1;
    if (kth <= sum(m))
      r = m;
    else
      l = m + 1;
  }
  return l;
}
bool _comp(const pair<long long, long long> &A,
           const pair<long long, long long> &B) {
  return A.first < B.first || (A.first == B.first && A.second > B.second);
}
bool __comp(const Query &A, const Query &B) { return A.k < B.k; }
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    long long x;
    cin >> x;
    arr[i] = x;
    vec.push_back({x, i});
  }
  sort(vec.begin(), vec.end(), _comp);
  aux.push_back(vec.back());
  add(aux.back().second);
  vec.pop_back();
  cin >> q;
  for (long long i = 0; i < q; ++i) {
    long long k, p;
    cin >> k >> p;
    query.push_back({k, p, i});
  }
  sort(query.begin(), query.end(), __comp);
  for (long long i = 0; i < query.size(); ++i) {
    while (query[i].k > aux.size()) {
      aux.push_back(vec.back());
      add(aux.back().second);
      vec.pop_back();
    }
    answer[query[i].index] = arr[very_quick_select(query[i].p)];
  }
  for (long long i = 0; i < q; ++i) {
    cout << answer[i] << "\n";
  }
  return 0;
}

```