#include <bits/stdc++.h>
using namespace std;
const int maxn = 2.3e5;
int tree[maxn * 4];
bool cmp_1(pair<int, int> &p_1, pair<int, int> &p_2) {
  if (p_1.first == p_2.first) {
    return p_1.second < p_2.second;
  } else {
    return p_1.first > p_2.first;
  }
}
void init() {
  for (int i = 0; i < maxn; ++i) {
    tree[i] = 0;
  }
}
void insert(int x, int node, int l, int r) {
  if (l == r) {
    tree[node]++;
    return;
  }
  int mid = (l + r) / 2;
  int left_node = 2 * node + 1;
  int right_node = left_node + 1;
  if (x >= l && x <= mid) {
    insert(x, left_node, l, mid);
  } else {
    insert(x, right_node, mid + 1, r);
  }
  tree[node] = tree[left_node] + tree[right_node];
}
int find_kth_num(int k, int node, int l, int r) {
  if (l == r) {
    return l;
  }
  int mid = (l + r) / 2;
  int left_node = 2 * node + 1;
  int right_node = left_node + 1;
  int l_cnt = tree[left_node];
  int r_cnt = tree[right_node];
  if (k > l_cnt) {
    k -= l_cnt;
    return find_kth_num(k, right_node, mid + 1, r);
  } else {
    return find_kth_num(k, left_node, l, mid);
  }
}
int main() {
  init();
  int n;
  cin >> n;
  vector<pair<int, int> > v;
  vector<int> v_ori;
  pair<int, int> p;
  for (int i = 0; i < n; ++i) {
    int num;
    cin >> num;
    p.first = num;
    p.second = i;
    v.push_back(p);
    v_ori.push_back(num);
  }
  sort(v.begin(), v.end(), cmp_1);
  map<int, vector<int> > mp;
  int m;
  cin >> m;
  vector<pair<int, int> > v_q;
  for (int i = 0; i < m; ++i) {
    int k, pos;
    cin >> k >> pos;
    mp[k].push_back(pos);
    p = make_pair(k, pos);
    v_q.push_back(p);
  }
  map<pair<int, int>, int> mp_re;
  for (int i = 0; i < n; ++i) {
    p = v[i];
    insert(p.second, 0, 0, n - 1);
    if (mp.count(i + 1)) {
      vector<int> q = mp[i + 1];
      int size = q.size();
      for (int j = 0; j < size; ++j) {
        int k = q[j];
        int index = find_kth_num(k, 0, 0, n - 1);
        int num = v_ori[index];
        mp_re[make_pair(i + 1, k)] = num;
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    p = v_q[i];
    cout << mp_re[p] << endl;
  }
  return 0;
}

### 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>
using namespace std;
const int maxn = 2.3e5;
int tree[maxn * 4];
bool cmp_1(pair<int, int> &p_1, pair<int, int> &p_2) {
  if (p_1.first == p_2.first) {
    return p_1.second < p_2.second;
  } else {
    return p_1.first > p_2.first;
  }
}
void init() {
  for (int i = 0; i < maxn; ++i) {
    tree[i] = 0;
  }
}
void insert(int x, int node, int l, int r) {
  if (l == r) {
    tree[node]++;
    return;
  }
  int mid = (l + r) / 2;
  int left_node = 2 * node + 1;
  int right_node = left_node + 1;
  if (x >= l && x <= mid) {
    insert(x, left_node, l, mid);
  } else {
    insert(x, right_node, mid + 1, r);
  }
  tree[node] = tree[left_node] + tree[right_node];
}
int find_kth_num(int k, int node, int l, int r) {
  if (l == r) {
    return l;
  }
  int mid = (l + r) / 2;
  int left_node = 2 * node + 1;
  int right_node = left_node + 1;
  int l_cnt = tree[left_node];
  int r_cnt = tree[right_node];
  if (k > l_cnt) {
    k -= l_cnt;
    return find_kth_num(k, right_node, mid + 1, r);
  } else {
    return find_kth_num(k, left_node, l, mid);
  }
}
int main() {
  init();
  int n;
  cin >> n;
  vector<pair<int, int> > v;
  vector<int> v_ori;
  pair<int, int> p;
  for (int i = 0; i < n; ++i) {
    int num;
    cin >> num;
    p.first = num;
    p.second = i;
    v.push_back(p);
    v_ori.push_back(num);
  }
  sort(v.begin(), v.end(), cmp_1);
  map<int, vector<int> > mp;
  int m;
  cin >> m;
  vector<pair<int, int> > v_q;
  for (int i = 0; i < m; ++i) {
    int k, pos;
    cin >> k >> pos;
    mp[k].push_back(pos);
    p = make_pair(k, pos);
    v_q.push_back(p);
  }
  map<pair<int, int>, int> mp_re;
  for (int i = 0; i < n; ++i) {
    p = v[i];
    insert(p.second, 0, 0, n - 1);
    if (mp.count(i + 1)) {
      vector<int> q = mp[i + 1];
      int size = q.size();
      for (int j = 0; j < size; ++j) {
        int k = q[j];
        int index = find_kth_num(k, 0, 0, n - 1);
        int num = v_ori[index];
        mp_re[make_pair(i + 1, k)] = num;
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    p = v_q[i];
    cout << mp_re[p] << endl;
  }
  return 0;
}

```