#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
int n, m;
int a[N];
int order[N];
int root[N], idx;
struct Node {
  int l, r, cnt;
} tr[N * 4 + N * 17];
void pushup(int u) { tr[u].cnt = tr[tr[u].l].cnt + tr[tr[u].r].cnt; }
int build(int l, int r) {
  int p = ++idx;
  if (l == r)
    return p;
  else {
    int mid = l + r >> 1;
    tr[p].l = build(l, mid);
    tr[p].r = build(mid + 1, r);
    return p;
  }
}
int insert(int p, int l, int r, int x) {
  int q = ++idx;
  tr[q] = tr[p];
  if (l == r) {
    tr[q].cnt++;
    return q;
  } else {
    int mid = l + r >> 1;
    if (x <= mid)
      tr[q].l = insert(tr[q].l, l, mid, x);
    else
      tr[q].r = insert(tr[q].r, mid + 1, r, x);
    pushup(q);
    return q;
  }
}
int query(int p, int q, int l, int r, int k) {
  if (l == r)
    return r;
  else {
    int cnt = tr[tr[q].l].cnt - tr[tr[p].l].cnt;
    int mid = l + r >> 1;
    if (cnt >= k)
      return query(tr[p].l, tr[q].l, l, mid, k);
    else
      return query(tr[p].r, tr[q].r, mid + 1, r, k - cnt);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    order[i] = i;
  }
  sort(order + 1, order + 1 + n, [&](int x, int y) {
    if (a[x] != a[y])
      return a[x] > a[y];
    else
      return x < y;
  });
  root[0] = build(1, n);
  for (int i = 1; i <= n; i++) root[i] = insert(root[i - 1], 1, n, order[i]);
  cin >> m;
  while (m--) {
    int k, pos;
    cin >> k >> pos;
    cout << a[query(root[0], root[k], 1, n, pos)] << '\n';
  }
  return 0;
}

### Prompt

Please formulate 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;
const int N = 200010;
int n, m;
int a[N];
int order[N];
int root[N], idx;
struct Node {
  int l, r, cnt;
} tr[N * 4 + N * 17];
void pushup(int u) { tr[u].cnt = tr[tr[u].l].cnt + tr[tr[u].r].cnt; }
int build(int l, int r) {
  int p = ++idx;
  if (l == r)
    return p;
  else {
    int mid = l + r >> 1;
    tr[p].l = build(l, mid);
    tr[p].r = build(mid + 1, r);
    return p;
  }
}
int insert(int p, int l, int r, int x) {
  int q = ++idx;
  tr[q] = tr[p];
  if (l == r) {
    tr[q].cnt++;
    return q;
  } else {
    int mid = l + r >> 1;
    if (x <= mid)
      tr[q].l = insert(tr[q].l, l, mid, x);
    else
      tr[q].r = insert(tr[q].r, mid + 1, r, x);
    pushup(q);
    return q;
  }
}
int query(int p, int q, int l, int r, int k) {
  if (l == r)
    return r;
  else {
    int cnt = tr[tr[q].l].cnt - tr[tr[p].l].cnt;
    int mid = l + r >> 1;
    if (cnt >= k)
      return query(tr[p].l, tr[q].l, l, mid, k);
    else
      return query(tr[p].r, tr[q].r, mid + 1, r, k - cnt);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    order[i] = i;
  }
  sort(order + 1, order + 1 + n, [&](int x, int y) {
    if (a[x] != a[y])
      return a[x] > a[y];
    else
      return x < y;
  });
  root[0] = build(1, n);
  for (int i = 1; i <= n; i++) root[i] = insert(root[i - 1], 1, n, order[i]);
  cin >> m;
  while (m--) {
    int k, pos;
    cin >> k >> pos;
    cout << a[query(root[0], root[k], 1, n, pos)] << '\n';
  }
  return 0;
}

```