#include <bits/stdc++.h>
using namespace std;
#pragma GCC diagnostic ignored "-Wmissing-declarations"
inline int safe_mul(const int x, const int y)
    __attribute__((warn_unused_result));
int const maxn = 200005;
int tree[maxn * 4];
int sz = 1;
int get(int need) {
  int idx = 1;
  while (idx < sz) {
    if (tree[2 * idx] < need) {
      need -= tree[2 * idx];
      idx = 2 * idx + 1;
    } else {
      idx = 2 * idx;
    }
  }
  return idx - sz;
}
inline void upd(int idx) {
  for (tree[idx += sz] += 1; idx > 1; idx >>= 1)
    tree[idx >> 1] = tree[idx] + tree[idx ^ 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> a;
  vector<int> nums;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    a.emplace_back(x, -i);
    nums.push_back(x);
  }
  sort(begin(a), end(a), greater<pair<int, int>>());
  while (sz <= n) sz <<= 1;
  int m;
  cin >> m;
  vector<tuple<int, int, int>> qu;
  for (int i = 0; i < m; ++i) {
    int k, p;
    cin >> k >> p;
    qu.emplace_back(k, p, i);
  }
  sort(begin(qu), end(qu));
  vector<int> ans(m);
  auto it = qu.begin();
  for (int i = 1; i <= n; ++i) {
    upd(-a[i - 1].second);
    while (it != qu.end() && get<0>(*it) == i) {
      int p, idx;
      tie(ignore, p, idx) = *it;
      ans[idx] = nums[get(p) - 1];
      ++it;
    }
  }
  for (int x : ans) cout << x << '\n';
}

### 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;
#pragma GCC diagnostic ignored "-Wmissing-declarations"
inline int safe_mul(const int x, const int y)
    __attribute__((warn_unused_result));
int const maxn = 200005;
int tree[maxn * 4];
int sz = 1;
int get(int need) {
  int idx = 1;
  while (idx < sz) {
    if (tree[2 * idx] < need) {
      need -= tree[2 * idx];
      idx = 2 * idx + 1;
    } else {
      idx = 2 * idx;
    }
  }
  return idx - sz;
}
inline void upd(int idx) {
  for (tree[idx += sz] += 1; idx > 1; idx >>= 1)
    tree[idx >> 1] = tree[idx] + tree[idx ^ 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> a;
  vector<int> nums;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    a.emplace_back(x, -i);
    nums.push_back(x);
  }
  sort(begin(a), end(a), greater<pair<int, int>>());
  while (sz <= n) sz <<= 1;
  int m;
  cin >> m;
  vector<tuple<int, int, int>> qu;
  for (int i = 0; i < m; ++i) {
    int k, p;
    cin >> k >> p;
    qu.emplace_back(k, p, i);
  }
  sort(begin(qu), end(qu));
  vector<int> ans(m);
  auto it = qu.begin();
  for (int i = 1; i <= n; ++i) {
    upd(-a[i - 1].second);
    while (it != qu.end() && get<0>(*it) == i) {
      int p, idx;
      tie(ignore, p, idx) = *it;
      ans[idx] = nums[get(p) - 1];
      ++it;
    }
  }
  for (int x : ans) cout << x << '\n';
}

```