#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 30;
const long long MOD = 1e9 + 7;
const long long MAX = 2e5 + 5;
const long long MAXX = 1e5 + 5;
long long powerm(long long a, long long b, long long m) {
  if (b == 0) {
    return 1;
  }
  long long res = powerm(a, b / 2, m);
  if (b % 2) {
    return ((res * res) % m * (a % m)) % m;
  }
  return (res * res) % m;
}
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long res = power(a, b / 2);
  if (b % 2 == 0) {
    return res * res;
  }
  return res * res * a;
}
long long b[MAX];
void add(long long a, long long x) {
  long long curr = a;
  while (curr < MAX) {
    b[curr] += x;
    curr += curr - (curr & (curr - 1));
  }
  return;
}
long long q(long long r) {
  long long curr = r, res = 0;
  while (curr > 0) {
    res += b[curr];
    curr &= (curr - 1);
  }
  return res;
}
bool cmp(pair<long long, long long>& a, pair<long long, long long>& b) {
  if (a.first == b.first) {
    return a.second < b.second;
  }
  return a.first > b.first;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n;
  long long yo[n];
  pair<long long, long long> arr[n];
  for (long long i = 0; i < n; i++) {
    cin >> arr[i].first;
    yo[i] = arr[i].first;
    arr[i].second = i + 1;
  }
  sort(arr, arr + n, cmp);
  cin >> m;
  long long ans[m];
  tuple<long long, long long, long long> r[m];
  for (long long i = 0; i < m; i++) {
    cin >> get<0>(r[i]) >> get<1>(r[i]);
    get<2>(r[i]) = i;
  }
  sort(r, r + m);
  long long curr = 0;
  for (long long i = 0; i < m; i++) {
    while (curr < get<0>(r[i])) {
      add(arr[curr].second, 1);
      curr++;
    }
    long long first = 1, last = n, res = -1;
    while (first <= last) {
      long long mid = (first + last) / 2;
      if (q(mid) >= get<1>(r[i])) {
        res = mid;
        last = mid - 1;
      } else {
        first = mid + 1;
      }
    }
    ans[get<2>(r[i])] = yo[res - 1];
  }
  for (long long i = 0; i < m; i++) {
    cout << ans[i] << "\n";
  }
  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 long long MAXN = 30;
const long long MOD = 1e9 + 7;
const long long MAX = 2e5 + 5;
const long long MAXX = 1e5 + 5;
long long powerm(long long a, long long b, long long m) {
  if (b == 0) {
    return 1;
  }
  long long res = powerm(a, b / 2, m);
  if (b % 2) {
    return ((res * res) % m * (a % m)) % m;
  }
  return (res * res) % m;
}
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long res = power(a, b / 2);
  if (b % 2 == 0) {
    return res * res;
  }
  return res * res * a;
}
long long b[MAX];
void add(long long a, long long x) {
  long long curr = a;
  while (curr < MAX) {
    b[curr] += x;
    curr += curr - (curr & (curr - 1));
  }
  return;
}
long long q(long long r) {
  long long curr = r, res = 0;
  while (curr > 0) {
    res += b[curr];
    curr &= (curr - 1);
  }
  return res;
}
bool cmp(pair<long long, long long>& a, pair<long long, long long>& b) {
  if (a.first == b.first) {
    return a.second < b.second;
  }
  return a.first > b.first;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n;
  long long yo[n];
  pair<long long, long long> arr[n];
  for (long long i = 0; i < n; i++) {
    cin >> arr[i].first;
    yo[i] = arr[i].first;
    arr[i].second = i + 1;
  }
  sort(arr, arr + n, cmp);
  cin >> m;
  long long ans[m];
  tuple<long long, long long, long long> r[m];
  for (long long i = 0; i < m; i++) {
    cin >> get<0>(r[i]) >> get<1>(r[i]);
    get<2>(r[i]) = i;
  }
  sort(r, r + m);
  long long curr = 0;
  for (long long i = 0; i < m; i++) {
    while (curr < get<0>(r[i])) {
      add(arr[curr].second, 1);
      curr++;
    }
    long long first = 1, last = n, res = -1;
    while (first <= last) {
      long long mid = (first + last) / 2;
      if (q(mid) >= get<1>(r[i])) {
        res = mid;
        last = mid - 1;
      } else {
        first = mid + 1;
      }
    }
    ans[get<2>(r[i])] = yo[res - 1];
  }
  for (long long i = 0; i < m; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}

```