#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long fp(long long a, long long b) {
  if (b == 0) return 1;
  long long x = fp(a, b / 2);
  x = (x * x) % mod;
  if (b & 1) x = (x * a) % mod;
  return x;
}
const long long N = 2e5 + 5;
class fenwick {
 public:
  vector<long long> bit;
  vector<long long> data;
  fenwick() {
    data = vector<long long>(N);
    bit = vector<long long>(N);
  }
  void add(long long node, long long x) {
    data[node] += x;
    for (; node < N; node += (node & (-node))) bit[node] += x;
  }
  long long get(long long k) {
    long long x = 0;
    for (long long i = log2(N); i >= 0; i--) {
      if ((x + (1ll << i) < N) && bit[x + (1ll << i)] < k) {
        k -= bit[x + (1ll << i)];
        x += (1ll << i);
      }
    }
    return x + 1;
  }
} f;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, x;
  cin >> n;
  map<long long, long long> mp;
  vector<array<long long, 2>> v;
  long long a[n + 1];
  long long b[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    mp[a[i]];
    b[i] = a[i];
  }
  long long ct = 0;
  for (auto &i : mp) i.second = ++ct;
  for (long long i = 1; i <= n; i++) b[i] = mp[b[i]];
  for (long long i = 1; i <= n; i++) {
    v.push_back({b[i], -i});
  }
  sort(v.rbegin(), v.rend());
  long long q;
  cin >> q;
  vector<array<long long, 2>> qr[n + 1];
  long long ans[q + 1], k, pos;
  for (long long i = 0; i < q; i++) {
    cin >> k >> pos;
    qr[k - 1].push_back({pos, i});
  }
  for (long long i = 0; i < v.size(); i++) {
    f.add(-v[i][1], 1);
    for (auto j : qr[i]) {
      ans[j[1]] = a[f.get(j[0])];
    }
  }
  for (long long i = 0; i < q; i++) cout << ans[i] << "\n";
}

### 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;
long long mod = 1e9 + 7;
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long fp(long long a, long long b) {
  if (b == 0) return 1;
  long long x = fp(a, b / 2);
  x = (x * x) % mod;
  if (b & 1) x = (x * a) % mod;
  return x;
}
const long long N = 2e5 + 5;
class fenwick {
 public:
  vector<long long> bit;
  vector<long long> data;
  fenwick() {
    data = vector<long long>(N);
    bit = vector<long long>(N);
  }
  void add(long long node, long long x) {
    data[node] += x;
    for (; node < N; node += (node & (-node))) bit[node] += x;
  }
  long long get(long long k) {
    long long x = 0;
    for (long long i = log2(N); i >= 0; i--) {
      if ((x + (1ll << i) < N) && bit[x + (1ll << i)] < k) {
        k -= bit[x + (1ll << i)];
        x += (1ll << i);
      }
    }
    return x + 1;
  }
} f;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, x;
  cin >> n;
  map<long long, long long> mp;
  vector<array<long long, 2>> v;
  long long a[n + 1];
  long long b[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    mp[a[i]];
    b[i] = a[i];
  }
  long long ct = 0;
  for (auto &i : mp) i.second = ++ct;
  for (long long i = 1; i <= n; i++) b[i] = mp[b[i]];
  for (long long i = 1; i <= n; i++) {
    v.push_back({b[i], -i});
  }
  sort(v.rbegin(), v.rend());
  long long q;
  cin >> q;
  vector<array<long long, 2>> qr[n + 1];
  long long ans[q + 1], k, pos;
  for (long long i = 0; i < q; i++) {
    cin >> k >> pos;
    qr[k - 1].push_back({pos, i});
  }
  for (long long i = 0; i < v.size(); i++) {
    f.add(-v[i][1], 1);
    for (auto j : qr[i]) {
      ans[j[1]] = a[f.get(j[0])];
    }
  }
  for (long long i = 0; i < q; i++) cout << ans[i] << "\n";
}

```