#include <bits/stdc++.h>
const double PI = 3.141592653589793238460;
using namespace std;
pair<long long int, long long int> ar[200001];
pair<long long int, long long int> st[800004];
pair<int, pair<long long int, long long int> > Q[200005];
int get(int si, int second, int se, int qi) {
  if (second == se) {
    return st[si].first;
  }
  int mid = (second + se) / 2;
  if (st[2 * si].second >= qi) {
    return get(2 * si, second, mid, qi);
  } else {
    return get(2 * si + 1, mid + 1, se, qi - st[2 * si].second);
  }
}
void update(int si, int second, int se, int qi, int val) {
  if (second == se) {
    st[si] = {val, 1};
    return;
  }
  int mid = (second + se) / 2;
  if (qi <= mid)
    update(2 * si, second, mid, qi, val);
  else
    update(2 * si + 1, mid + 1, se, qi, val);
  st[si].second = st[2 * si].second + st[2 * si + 1].second;
}
bool comp(pair<long long int, long long int> a,
          pair<long long int, long long int> b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second < b.second;
}
int ans[200001];
int main() {
  int n, pos, k, q;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> ar[i].first, ar[i].second = i;
  sort(ar + 1, ar + n + 1, comp);
  cin >> q;
  for (int i = 1; i <= q; i++)
    cin >> Q[i].first >> Q[i].second.first, Q[i].second.second = i;
  sort(Q + 1, Q + q + 1);
  int j = 0;
  for (int i = 1; i <= q; i++) {
    k = Q[i].first;
    pos = Q[i].second.first;
    while (j < k) update(1, 1, n, ar[j + 1].second, ar[j + 1].first), j++;
    ans[Q[i].second.second] = get(1, 1, n, pos);
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << '\n';
}

### Prompt

Please provide a CPP coded solution 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>
const double PI = 3.141592653589793238460;
using namespace std;
pair<long long int, long long int> ar[200001];
pair<long long int, long long int> st[800004];
pair<int, pair<long long int, long long int> > Q[200005];
int get(int si, int second, int se, int qi) {
  if (second == se) {
    return st[si].first;
  }
  int mid = (second + se) / 2;
  if (st[2 * si].second >= qi) {
    return get(2 * si, second, mid, qi);
  } else {
    return get(2 * si + 1, mid + 1, se, qi - st[2 * si].second);
  }
}
void update(int si, int second, int se, int qi, int val) {
  if (second == se) {
    st[si] = {val, 1};
    return;
  }
  int mid = (second + se) / 2;
  if (qi <= mid)
    update(2 * si, second, mid, qi, val);
  else
    update(2 * si + 1, mid + 1, se, qi, val);
  st[si].second = st[2 * si].second + st[2 * si + 1].second;
}
bool comp(pair<long long int, long long int> a,
          pair<long long int, long long int> b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second < b.second;
}
int ans[200001];
int main() {
  int n, pos, k, q;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> ar[i].first, ar[i].second = i;
  sort(ar + 1, ar + n + 1, comp);
  cin >> q;
  for (int i = 1; i <= q; i++)
    cin >> Q[i].first >> Q[i].second.first, Q[i].second.second = i;
  sort(Q + 1, Q + q + 1);
  int j = 0;
  for (int i = 1; i <= q; i++) {
    k = Q[i].first;
    pos = Q[i].second.first;
    while (j < k) update(1, 1, n, ar[j + 1].second, ar[j + 1].first), j++;
    ans[Q[i].second.second] = get(1, 1, n, pos);
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << '\n';
}

```