#include <bits/stdc++.h>
using namespace std;
int seg[(int)(2e5 + 1) * 4];
void update(int s, int e, int idx, int k) {
  if (s == e && s == k) {
    seg[idx] = 1;
    return;
  }
  if (s > k || e < k) return;
  int mid = (s + e) / 2;
  update(s, mid, idx * 2, k);
  update(mid + 1, e, idx * 2 + 1, k);
  seg[idx] = seg[idx * 2] + seg[idx * 2 + 1];
}
int kth(int s, int e, int idx, int k) {
  if (s == e) return s;
  int mid = (s + e) / 2;
  if (seg[idx * 2] >= k)
    return kth(s, mid, idx * 2, k);
  else
    return kth(mid + 1, e, idx * 2 + 1, k - seg[idx * 2]);
}
void look(int s, int e, int idx) {
  if (s == e) {
    return;
  }
  int mid = (s + e) / 2;
  look(s, mid, idx * 2);
  look(mid + 1, e, idx * 2 + 1);
}
struct ano {
  int k;
  int pos;
  int id;
};
bool cmp(ano a, ano b) {
  if (a.k == b.k) return a.pos < b.pos;
  return a.k < b.k;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int i;
  int arr[n];
  pair<int, int> sorted[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    sorted[i].first = -arr[i];
    sorted[i].second = i;
  }
  sort(sorted, sorted + n);
  int q;
  cin >> q;
  ano ask[q];
  for (i = 0; i < q; i++) {
    cin >> ask[i].k >> ask[i].pos;
    ask[i].id = i;
  }
  sort(ask, ask + q, cmp);
  int ans[q];
  int curr = -1;
  for (i = 0; i < q; i++) {
    while (curr < ask[i].k - 1) {
      curr++;
      update(0, n - 1, 1, sorted[curr].second);
    }
    int pos = kth(0, n - 1, 1, ask[i].pos);
    ans[ask[i].id] = arr[pos];
  }
  for (i = 0; i < q; i++) cout << ans[i] << '\n';
}

### Prompt

Please formulate a cpp solution 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;
int seg[(int)(2e5 + 1) * 4];
void update(int s, int e, int idx, int k) {
  if (s == e && s == k) {
    seg[idx] = 1;
    return;
  }
  if (s > k || e < k) return;
  int mid = (s + e) / 2;
  update(s, mid, idx * 2, k);
  update(mid + 1, e, idx * 2 + 1, k);
  seg[idx] = seg[idx * 2] + seg[idx * 2 + 1];
}
int kth(int s, int e, int idx, int k) {
  if (s == e) return s;
  int mid = (s + e) / 2;
  if (seg[idx * 2] >= k)
    return kth(s, mid, idx * 2, k);
  else
    return kth(mid + 1, e, idx * 2 + 1, k - seg[idx * 2]);
}
void look(int s, int e, int idx) {
  if (s == e) {
    return;
  }
  int mid = (s + e) / 2;
  look(s, mid, idx * 2);
  look(mid + 1, e, idx * 2 + 1);
}
struct ano {
  int k;
  int pos;
  int id;
};
bool cmp(ano a, ano b) {
  if (a.k == b.k) return a.pos < b.pos;
  return a.k < b.k;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int i;
  int arr[n];
  pair<int, int> sorted[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    sorted[i].first = -arr[i];
    sorted[i].second = i;
  }
  sort(sorted, sorted + n);
  int q;
  cin >> q;
  ano ask[q];
  for (i = 0; i < q; i++) {
    cin >> ask[i].k >> ask[i].pos;
    ask[i].id = i;
  }
  sort(ask, ask + q, cmp);
  int ans[q];
  int curr = -1;
  for (i = 0; i < q; i++) {
    while (curr < ask[i].k - 1) {
      curr++;
      update(0, n - 1, 1, sorted[curr].second);
    }
    int pos = kth(0, n - 1, 1, ask[i].pos);
    ans[ask[i].id] = arr[pos];
  }
  for (i = 0; i < q; i++) cout << ans[i] << '\n';
}

```