#include <bits/stdc++.h>
using namespace std;
int A[(200000 + 4)], S[(200000 + 4)];
void add(int pos, int val) {
  while (pos < (200000 + 4)) {
    A[pos] += val;
    pos += ((pos) & (-(pos)));
  }
}
int sum(int x) {
  int Ans = 0;
  while (x) {
    Ans += A[x];
    x -= ((x) & (-(x)));
  }
  return Ans;
}
struct point {
  int pos, val;
  bool operator<(const point &b) const {
    return val == b.val ? pos < b.pos : val > b.val;
  }
} P[(200000 + 4)];
struct query {
  int cnt, pos, ans, id;
  bool operator<(const query &b) const { return cnt < b.cnt; }
} Q[(200000 + 4)];
bool cmp(query A, query B) { return A.id < B.id; }
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> P[i].val;
    S[i] = P[i].val;
    P[i].pos = i;
  }
  sort(P + 1, P + 1 + n);
  int m;
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> Q[i].cnt >> Q[i].pos;
    Q[i].id = i;
  }
  sort(Q + 1, Q + 1 + m);
  int size = 0;
  for (register int i = 1; i <= m; ++i) {
    while (size < Q[i].cnt) {
      ++size;
      add(P[size].pos, 1);
    }
    int l = 0, r = n + 1, mid = 0;
    while (l < r) {
      mid = l + r >> 1;
      if (sum(mid) >= Q[i].pos)
        r = mid;
      else
        l = mid + 1;
    }
    Q[i].ans = S[r];
  }
  sort(Q + 1, Q + 1 + m, cmp);
  for (int i = 1; i <= m; ++i) cout << Q[i].ans << "\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;
int A[(200000 + 4)], S[(200000 + 4)];
void add(int pos, int val) {
  while (pos < (200000 + 4)) {
    A[pos] += val;
    pos += ((pos) & (-(pos)));
  }
}
int sum(int x) {
  int Ans = 0;
  while (x) {
    Ans += A[x];
    x -= ((x) & (-(x)));
  }
  return Ans;
}
struct point {
  int pos, val;
  bool operator<(const point &b) const {
    return val == b.val ? pos < b.pos : val > b.val;
  }
} P[(200000 + 4)];
struct query {
  int cnt, pos, ans, id;
  bool operator<(const query &b) const { return cnt < b.cnt; }
} Q[(200000 + 4)];
bool cmp(query A, query B) { return A.id < B.id; }
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> P[i].val;
    S[i] = P[i].val;
    P[i].pos = i;
  }
  sort(P + 1, P + 1 + n);
  int m;
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> Q[i].cnt >> Q[i].pos;
    Q[i].id = i;
  }
  sort(Q + 1, Q + 1 + m);
  int size = 0;
  for (register int i = 1; i <= m; ++i) {
    while (size < Q[i].cnt) {
      ++size;
      add(P[size].pos, 1);
    }
    int l = 0, r = n + 1, mid = 0;
    while (l < r) {
      mid = l + r >> 1;
      if (sum(mid) >= Q[i].pos)
        r = mid;
      else
        l = mid + 1;
    }
    Q[i].ans = S[r];
  }
  sort(Q + 1, Q + 1 + m, cmp);
  for (int i = 1; i <= m; ++i) cout << Q[i].ans << "\n";
  return 0;
}

```