#include <bits/stdc++.h>
const long long int MOD = 1e9 + 7;
using namespace std;
vector<int> indx, tree[4 * 200005];
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  if (a.first > b.first)
    return true;
  else if (a.first == b.first) {
    if (a.second < b.second) return true;
    return false;
  } else
    return false;
}
void combine(int node) {
  merge(tree[2 * node].begin(), tree[2 * node].end(),
        tree[2 * node + 1].begin(), tree[2 * node + 1].end(),
        back_inserter(tree[node]));
}
void build(int node, int l, int r) {
  if (l == r) {
    tree[node].push_back(indx[l]);
  } else {
    int m = (l + r) / 2;
    build(2 * node, l, m);
    build(2 * node + 1, m + 1, r);
    combine(node);
  }
}
int query(int node, int st, int en, long long int val, int l, int r) {
  if (st > r || en < l) return 0;
  if (st >= l && en <= r) {
    return upper_bound(tree[node].begin(), tree[node].end(), val) -
           tree[node].begin();
  } else {
    int m = (st + en) / 2;
    return query(2 * node, st, m, val, l, r) +
           query(2 * node + 1, m + 1, en, val, l, r);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  long long int arr[n + 1];
  std::vector<pair<long long int, long long int> > v(n);
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
    v[i - 1] = make_pair(arr[i], i);
  }
  sort(v.begin(), v.end(), cmp);
  indx.resize(n + 1);
  for (int i = 1; i <= n; ++i) {
    indx[i] = v[i - 1].second;
  }
  build(1, 1, n);
  int m;
  cin >> m;
  while (m--) {
    int k, p;
    cin >> k >> p;
    int ans = -1, l = 0, h = 200005;
    while (l <= h) {
      int m = (l + h) / 2;
      int ltm = query(1, 1, n, m, 1, k);
      if (ltm >= p) {
        ans = m;
        h = m - 1;
      } else
        l = m + 1;
    }
    cout << arr[ans] << '\n';
  }
  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>
const long long int MOD = 1e9 + 7;
using namespace std;
vector<int> indx, tree[4 * 200005];
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  if (a.first > b.first)
    return true;
  else if (a.first == b.first) {
    if (a.second < b.second) return true;
    return false;
  } else
    return false;
}
void combine(int node) {
  merge(tree[2 * node].begin(), tree[2 * node].end(),
        tree[2 * node + 1].begin(), tree[2 * node + 1].end(),
        back_inserter(tree[node]));
}
void build(int node, int l, int r) {
  if (l == r) {
    tree[node].push_back(indx[l]);
  } else {
    int m = (l + r) / 2;
    build(2 * node, l, m);
    build(2 * node + 1, m + 1, r);
    combine(node);
  }
}
int query(int node, int st, int en, long long int val, int l, int r) {
  if (st > r || en < l) return 0;
  if (st >= l && en <= r) {
    return upper_bound(tree[node].begin(), tree[node].end(), val) -
           tree[node].begin();
  } else {
    int m = (st + en) / 2;
    return query(2 * node, st, m, val, l, r) +
           query(2 * node + 1, m + 1, en, val, l, r);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  long long int arr[n + 1];
  std::vector<pair<long long int, long long int> > v(n);
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
    v[i - 1] = make_pair(arr[i], i);
  }
  sort(v.begin(), v.end(), cmp);
  indx.resize(n + 1);
  for (int i = 1; i <= n; ++i) {
    indx[i] = v[i - 1].second;
  }
  build(1, 1, n);
  int m;
  cin >> m;
  while (m--) {
    int k, p;
    cin >> k >> p;
    int ans = -1, l = 0, h = 200005;
    while (l <= h) {
      int m = (l + h) / 2;
      int ltm = query(1, 1, n, m, 1, k);
      if (ltm >= p) {
        ans = m;
        h = m - 1;
      } else
        l = m + 1;
    }
    cout << arr[ans] << '\n';
  }
  return 0;
}

```