#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000;
int a[MAXN + 5], bit[MAXN + 5];
int lowbit(int x) { return x & -x; }
void add(int i, int x) {
  while (i <= MAXN) {
    bit[i] += x;
    i += lowbit(i);
  }
}
int sum(int i) {
  int ans = 0;
  while (i) {
    ans += bit[i];
    i -= lowbit(i);
  }
  return ans;
}
void compress(vector<int>& a, vector<int>& b) {
  vector<int> v(a.begin(), a.end());
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  int n = a.size();
  for (int i = 0; i < n; i++) {
    b[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  }
}
bool cmp(pair<int, int>& a, pair<int, int>& b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second < b.second;
}
void work(pair<int, int>& q, map<pair<int, int>, int>& mp) {
  int k = q.first, p = q.second;
  int l = 1, r = MAXN, res = -1;
  while (l <= r) {
    int m = l + r >> 1;
    if (sum(m) < p)
      l = m + 1;
    else {
      res = m;
      r = m - 1;
    }
  }
  assert(res != -1);
  mp[q] = res;
}
int main() {
  int n, m;
  while (cin >> n) {
    memset(bit, 0, sizeof bit);
    vector<int> a(n), new_a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    compress(a, new_a);
    vector<pair<int, int> > b(n);
    for (int i = 0; i < n; i++) {
      b[i].first = a[i];
      b[i].second = i + 1;
    }
    sort(b.begin(), b.end(), cmp);
    cin >> m;
    vector<pair<int, int> > q(m);
    for (int i = 0; i < m; i++) cin >> q[i].first >> q[i].second;
    vector<pair<int, int> > sorted_q(q.begin(), q.end());
    sort(sorted_q.begin(), sorted_q.end());
    map<pair<int, int>, int> res;
    int cur = 0;
    for (int i = 0; i < n; i++) {
      add(b[i].second, 1);
      while (cur < m && i + 1 == sorted_q[cur].first) {
        work(sorted_q[cur], res);
        cur++;
      }
    }
    for (auto it : q) {
      int id = res[it];
      cout << a[id - 1] << endl;
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int MAXN = 200000;
int a[MAXN + 5], bit[MAXN + 5];
int lowbit(int x) { return x & -x; }
void add(int i, int x) {
  while (i <= MAXN) {
    bit[i] += x;
    i += lowbit(i);
  }
}
int sum(int i) {
  int ans = 0;
  while (i) {
    ans += bit[i];
    i -= lowbit(i);
  }
  return ans;
}
void compress(vector<int>& a, vector<int>& b) {
  vector<int> v(a.begin(), a.end());
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  int n = a.size();
  for (int i = 0; i < n; i++) {
    b[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  }
}
bool cmp(pair<int, int>& a, pair<int, int>& b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second < b.second;
}
void work(pair<int, int>& q, map<pair<int, int>, int>& mp) {
  int k = q.first, p = q.second;
  int l = 1, r = MAXN, res = -1;
  while (l <= r) {
    int m = l + r >> 1;
    if (sum(m) < p)
      l = m + 1;
    else {
      res = m;
      r = m - 1;
    }
  }
  assert(res != -1);
  mp[q] = res;
}
int main() {
  int n, m;
  while (cin >> n) {
    memset(bit, 0, sizeof bit);
    vector<int> a(n), new_a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    compress(a, new_a);
    vector<pair<int, int> > b(n);
    for (int i = 0; i < n; i++) {
      b[i].first = a[i];
      b[i].second = i + 1;
    }
    sort(b.begin(), b.end(), cmp);
    cin >> m;
    vector<pair<int, int> > q(m);
    for (int i = 0; i < m; i++) cin >> q[i].first >> q[i].second;
    vector<pair<int, int> > sorted_q(q.begin(), q.end());
    sort(sorted_q.begin(), sorted_q.end());
    map<pair<int, int>, int> res;
    int cur = 0;
    for (int i = 0; i < n; i++) {
      add(b[i].second, 1);
      while (cur < m && i + 1 == sorted_q[cur].first) {
        work(sorted_q[cur], res);
        cur++;
      }
    }
    for (auto it : q) {
      int id = res[it];
      cout << a[id - 1] << endl;
    }
  }
  return 0;
}

```