#include <bits/stdc++.h>
using namespace std;
long long n, m, k, vis[200000], ans[200000], seg[800000], h = 1;
pair<long long, long long> arr[200000];
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
void upd(long long x) {
  seg[x] = seg[2 * x] + seg[2 * x + 1];
  if (x > 1) upd(x / 2);
}
long long get(long long ix, long long iy, long long nx, long long ny,
              long long idx) {
  if (iy < nx || ny < ix) return 0;
  if (ix <= nx && ny <= iy) return seg[idx];
  return get(ix, iy, nx, (nx + ny) / 2, idx * 2) +
         get(ix, iy, (nx + ny) / 2 + 1, ny, idx * 2 + 1);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  while (h < n) h <<= 1;
  for (long long i = (0); i < (n); i++) {
    long long x;
    cin >> x;
    arr[i] = {x, i};
  }
  sort(arr, arr + n, cmp);
  for (long long i = (0); i < (n); i++) vis[arr[i].second] = i;
  vector<pair<long long, pair<long long, long long> > > v;
  cin >> m;
  for (long long i = (0); i < (m); i++) {
    long long k, p;
    cin >> k >> p;
    v.push_back({k, {p, i}});
  }
  sort(v.begin(), v.end());
  long long a = 0, j = 0;
  for (long long i = (0); i < (m); i++) {
    for (; j < v[i].first; j++) {
      seg[h + arr[j].second] = 1;
      upd((h + arr[j].second) / 2);
    }
    long long lo = 0, hi = n - 1;
    while (lo < hi) {
      long long mid = (lo + hi) / 2;
      if (get(0, mid, 0, h - 1, 1) >= v[i].second.first)
        hi = mid;
      else
        lo = mid + 1;
    }
    ans[v[i].second.second] = arr[vis[lo]].first;
  }
  for (long long i = (0); i < (m); i++) cout << ans[i] << '\n';
}

### Prompt

Your task is to create 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;
long long n, m, k, vis[200000], ans[200000], seg[800000], h = 1;
pair<long long, long long> arr[200000];
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
void upd(long long x) {
  seg[x] = seg[2 * x] + seg[2 * x + 1];
  if (x > 1) upd(x / 2);
}
long long get(long long ix, long long iy, long long nx, long long ny,
              long long idx) {
  if (iy < nx || ny < ix) return 0;
  if (ix <= nx && ny <= iy) return seg[idx];
  return get(ix, iy, nx, (nx + ny) / 2, idx * 2) +
         get(ix, iy, (nx + ny) / 2 + 1, ny, idx * 2 + 1);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  while (h < n) h <<= 1;
  for (long long i = (0); i < (n); i++) {
    long long x;
    cin >> x;
    arr[i] = {x, i};
  }
  sort(arr, arr + n, cmp);
  for (long long i = (0); i < (n); i++) vis[arr[i].second] = i;
  vector<pair<long long, pair<long long, long long> > > v;
  cin >> m;
  for (long long i = (0); i < (m); i++) {
    long long k, p;
    cin >> k >> p;
    v.push_back({k, {p, i}});
  }
  sort(v.begin(), v.end());
  long long a = 0, j = 0;
  for (long long i = (0); i < (m); i++) {
    for (; j < v[i].first; j++) {
      seg[h + arr[j].second] = 1;
      upd((h + arr[j].second) / 2);
    }
    long long lo = 0, hi = n - 1;
    while (lo < hi) {
      long long mid = (lo + hi) / 2;
      if (get(0, mid, 0, h - 1, 1) >= v[i].second.first)
        hi = mid;
      else
        lo = mid + 1;
    }
    ans[v[i].second.second] = arr[vis[lo]].first;
  }
  for (long long i = (0); i < (m); i++) cout << ans[i] << '\n';
}

```