#include <bits/stdc++.h>
using namespace std;
void read(int &x) { scanf("%d", &x); }
void read(long long int &x) { scanf("%I64d", &x); }
void read(long long int &a, long long int &b) { scanf("%I64d %I64d", &a, &b); }
void read(long long int &a, long long int &b, long long int &c) {
  scanf("%I64d %I64d %I64d", &a, &b, &c);
}
void read(long long int &a, long long int &b, long long int &c,
          long long int &d) {
  scanf("%I64d %I64d %I64d %I64d", &a, &b, &c, &d);
}
void read(double &x) { scanf("%lf", &x); }
void read(char &x) { scanf("%c", &x); }
void read(string &x) {
  char y[1000006];
  scanf("%s", &y);
  x = y;
}
void print(long long int x, bool flag = false) {
  if (flag)
    printf("%I64d ", x);
  else
    printf("%I64d\n", x);
}
void print(int x, bool flag = false) {
  if (flag)
    printf("%d ", x);
  else
    printf("%d\n", x);
}
void print(double x, bool flag = false) {
  if (flag)
    printf("%0.14lf ", x);
  else
    printf("%0.14lf\n", x);
}
void print(char x) { printf("%c\n", x); }
void print(string x) {
  for (long long int i = 0; i < x.size(); i++) {
    printf("%c", x[i]);
  }
  printf("\n");
}
const long long int mx = 2e5 + 2;
struct queries {
  long long int k, pos, id;
} q[mx];
bool comp(queries q1, queries q2) { return q1.k > q2.k; }
long long int fen[mx];
void update(long long int id, long long int val) {
  while (id < mx) {
    fen[id] += val;
    id += (id & -id);
  }
}
long long int query(long long int id) {
  long long int res = 0;
  while (id > 0) {
    res += fen[id];
    id -= (id & -id);
  }
  return res;
}
long long int find_index(long long int pos) {
  long long int lo = 1, high = mx - 2, ans = mx - 2;
  while (lo <= high) {
    long long int mid = (lo + high) / 2;
    if (query(mid) >= pos)
      ans = mid, high = mid - 1;
    else
      lo = mid + 1;
  }
  return ans;
}
void solve() {
  long long int n;
  read(n);
  long long int a[n + 1], cnt = 0;
  unordered_map<long long int, long long int> mp;
  multiset<long long int> s;
  long long int posi = n;
  for (long long int i = 1; i <= n; i++) {
    read(a[i]);
    s.insert(a[i]);
    update(i, 1);
  }
  for (auto p : s) {
    if (mp[p] == 0) mp[p] = ++cnt;
  }
  vector<long long int> index[n + 1];
  for (long long int i = 1; i <= n; i++) index[mp[a[i]]].push_back(i);
  long long int m;
  read(m);
  for (long long int i = 0; i < m; i++) {
    long long int k, pos;
    read(k, pos);
    q[i].k = k;
    q[i].pos = pos;
    q[i].id = i + 1;
  }
  long long int ans[m + 1];
  sort(q, q + m, comp);
  for (long long int i = 0; i < m; i++) {
    while (q[i].k < posi) {
      posi--;
      long long int mini = *s.begin();
      s.erase(s.begin());
      long long int ind = index[mp[mini]].back();
      index[mp[mini]].pop_back();
      update(ind, -1);
    }
    long long int ind = find_index(q[i].pos);
    ans[q[i].id] = a[ind];
  }
  for (long long int i = 1; i <= m; i++) print(ans[i]);
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
void read(int &x) { scanf("%d", &x); }
void read(long long int &x) { scanf("%I64d", &x); }
void read(long long int &a, long long int &b) { scanf("%I64d %I64d", &a, &b); }
void read(long long int &a, long long int &b, long long int &c) {
  scanf("%I64d %I64d %I64d", &a, &b, &c);
}
void read(long long int &a, long long int &b, long long int &c,
          long long int &d) {
  scanf("%I64d %I64d %I64d %I64d", &a, &b, &c, &d);
}
void read(double &x) { scanf("%lf", &x); }
void read(char &x) { scanf("%c", &x); }
void read(string &x) {
  char y[1000006];
  scanf("%s", &y);
  x = y;
}
void print(long long int x, bool flag = false) {
  if (flag)
    printf("%I64d ", x);
  else
    printf("%I64d\n", x);
}
void print(int x, bool flag = false) {
  if (flag)
    printf("%d ", x);
  else
    printf("%d\n", x);
}
void print(double x, bool flag = false) {
  if (flag)
    printf("%0.14lf ", x);
  else
    printf("%0.14lf\n", x);
}
void print(char x) { printf("%c\n", x); }
void print(string x) {
  for (long long int i = 0; i < x.size(); i++) {
    printf("%c", x[i]);
  }
  printf("\n");
}
const long long int mx = 2e5 + 2;
struct queries {
  long long int k, pos, id;
} q[mx];
bool comp(queries q1, queries q2) { return q1.k > q2.k; }
long long int fen[mx];
void update(long long int id, long long int val) {
  while (id < mx) {
    fen[id] += val;
    id += (id & -id);
  }
}
long long int query(long long int id) {
  long long int res = 0;
  while (id > 0) {
    res += fen[id];
    id -= (id & -id);
  }
  return res;
}
long long int find_index(long long int pos) {
  long long int lo = 1, high = mx - 2, ans = mx - 2;
  while (lo <= high) {
    long long int mid = (lo + high) / 2;
    if (query(mid) >= pos)
      ans = mid, high = mid - 1;
    else
      lo = mid + 1;
  }
  return ans;
}
void solve() {
  long long int n;
  read(n);
  long long int a[n + 1], cnt = 0;
  unordered_map<long long int, long long int> mp;
  multiset<long long int> s;
  long long int posi = n;
  for (long long int i = 1; i <= n; i++) {
    read(a[i]);
    s.insert(a[i]);
    update(i, 1);
  }
  for (auto p : s) {
    if (mp[p] == 0) mp[p] = ++cnt;
  }
  vector<long long int> index[n + 1];
  for (long long int i = 1; i <= n; i++) index[mp[a[i]]].push_back(i);
  long long int m;
  read(m);
  for (long long int i = 0; i < m; i++) {
    long long int k, pos;
    read(k, pos);
    q[i].k = k;
    q[i].pos = pos;
    q[i].id = i + 1;
  }
  long long int ans[m + 1];
  sort(q, q + m, comp);
  for (long long int i = 0; i < m; i++) {
    while (q[i].k < posi) {
      posi--;
      long long int mini = *s.begin();
      s.erase(s.begin());
      long long int ind = index[mp[mini]].back();
      index[mp[mini]].pop_back();
      update(ind, -1);
    }
    long long int ind = find_index(q[i].pos);
    ans[q[i].id] = a[ind];
  }
  for (long long int i = 1; i <= m; i++) print(ans[i]);
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```