#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<int, int>& a, pair<int, int>& b) {
  if (a.first == b.first) return a.second > b.second;
  return a.first < b.first;
}
const int NN = 2 * 1e5 + 3;
int a[NN], t[4 * NN];
void build(int v, int l, int r) {
  if (l == r - 1) {
    t[v] = a[l];
    return;
  }
  int m = (l + r) / 2;
  build(2 * v + 1, l, m);
  build(2 * v + 2, m, r);
  t[v] = t[2 * v + 1] + t[2 * v + 2];
}
int fnd(int v, int l, int r, int k) {
  if (k > t[v]) return -1;
  if (l == r - 1) return l;
  int m = (l + r) / 2;
  if (t[2 * v + 1] >= k)
    return fnd(2 * v + 1, l, m, k);
  else
    return fnd(2 * v + 2, m, r, k - t[2 * v + 1]);
}
void upd(int v, int l, int r, int id, int val) {
  if (l == r - 1) {
    t[v] = val;
    a[l] = val;
    return;
  }
  int m = (l + r) / 2;
  if (id < m)
    upd(2 * v + 1, l, m, id, val);
  else
    upd(2 * v + 2, m, r, id, val);
  t[v] = t[2 * v + 1] + t[2 * v + 2];
}
int sum(int v, int l, int r, int lf, int rg) {
  if (l == lf && r == rg) return t[v];
  int m = (l + r) / 2;
  int ans = 0;
  if (lf < m) ans += sum(2 * v + 1, l, m, lf, min(m, rg));
  if (rg > m) ans += sum(2 * v + 2, m, r, max(m, lf), rg);
  return ans;
}
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i];
  vector<pair<int, int>> del(n);
  for (int i = 0; i < n; ++i) del[i] = {v[i], i};
  sort(del.begin(), del.end(), cmp);
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int>> vp;
  for (int i = 0; i < m; i++) {
    int k, pos;
    cin >> k >> pos;
    vp.push_back({{k, pos}, i});
  }
  sort(vp.begin(), vp.end(),
       [](pair<pair<int, int>, int> v1, pair<pair<int, int>, int> v2) {
         if (v1.first.first == v2.first.first &&
             v1.first.second == v2.first.second)
           return v1.second < v2.second;
         else if (v1.first.first == v2.first.first)
           return v1.first.second < v2.first.second;
         else
           return v1.first.first > v2.first.first;
       });
  for (int i = 0; i < n; ++i) a[i] = 1;
  build(0, 0, n);
  int j = 0;
  int cur = n;
  vector<int> ans(m);
  for (int i = 0; i < m; ++i) {
    int kk = vp[i].first.first, poss = vp[i].first.second, id = vp[i].second;
    if (kk < cur) {
      while (cur > kk) {
        upd(0, 0, n, del[j].second, 0);
        j++;
        cur--;
      }
    }
    ans[id] = v[fnd(0, 0, n, poss)];
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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;
bool cmp(pair<int, int>& a, pair<int, int>& b) {
  if (a.first == b.first) return a.second > b.second;
  return a.first < b.first;
}
const int NN = 2 * 1e5 + 3;
int a[NN], t[4 * NN];
void build(int v, int l, int r) {
  if (l == r - 1) {
    t[v] = a[l];
    return;
  }
  int m = (l + r) / 2;
  build(2 * v + 1, l, m);
  build(2 * v + 2, m, r);
  t[v] = t[2 * v + 1] + t[2 * v + 2];
}
int fnd(int v, int l, int r, int k) {
  if (k > t[v]) return -1;
  if (l == r - 1) return l;
  int m = (l + r) / 2;
  if (t[2 * v + 1] >= k)
    return fnd(2 * v + 1, l, m, k);
  else
    return fnd(2 * v + 2, m, r, k - t[2 * v + 1]);
}
void upd(int v, int l, int r, int id, int val) {
  if (l == r - 1) {
    t[v] = val;
    a[l] = val;
    return;
  }
  int m = (l + r) / 2;
  if (id < m)
    upd(2 * v + 1, l, m, id, val);
  else
    upd(2 * v + 2, m, r, id, val);
  t[v] = t[2 * v + 1] + t[2 * v + 2];
}
int sum(int v, int l, int r, int lf, int rg) {
  if (l == lf && r == rg) return t[v];
  int m = (l + r) / 2;
  int ans = 0;
  if (lf < m) ans += sum(2 * v + 1, l, m, lf, min(m, rg));
  if (rg > m) ans += sum(2 * v + 2, m, r, max(m, lf), rg);
  return ans;
}
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i];
  vector<pair<int, int>> del(n);
  for (int i = 0; i < n; ++i) del[i] = {v[i], i};
  sort(del.begin(), del.end(), cmp);
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int>> vp;
  for (int i = 0; i < m; i++) {
    int k, pos;
    cin >> k >> pos;
    vp.push_back({{k, pos}, i});
  }
  sort(vp.begin(), vp.end(),
       [](pair<pair<int, int>, int> v1, pair<pair<int, int>, int> v2) {
         if (v1.first.first == v2.first.first &&
             v1.first.second == v2.first.second)
           return v1.second < v2.second;
         else if (v1.first.first == v2.first.first)
           return v1.first.second < v2.first.second;
         else
           return v1.first.first > v2.first.first;
       });
  for (int i = 0; i < n; ++i) a[i] = 1;
  build(0, 0, n);
  int j = 0;
  int cur = n;
  vector<int> ans(m);
  for (int i = 0; i < m; ++i) {
    int kk = vp[i].first.first, poss = vp[i].first.second, id = vp[i].second;
    if (kk < cur) {
      while (cur > kk) {
        upd(0, 0, n, del[j].second, 0);
        j++;
        cur--;
      }
    }
    ans[id] = v[fnd(0, 0, n, poss)];
  }
  for (int i = 0; i < m; ++i) cout << ans[i] << endl;
  return 0;
}

```