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]. 
#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a < 0 ? a + P : a;
}
int mul(int a, int b) { return 1ll * a * b % P; }
int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 2e5 + 5;
int n, m;
pair<int, int> a[N];
struct node {
  int pos, k, id;
} q[N];
bool cmp(node a, node b) { return a.k < b.k; }
int ans[N], b[N << 2], c[N];
void upd(int L, int l, int r, int rt) {
  if (l == r) {
    b[rt] = 1;
    return;
  }
  int m = l + r >> 1;
  if (L <= m)
    upd(L, l, m, rt << 1);
  else
    upd(L, m + 1, r, rt << 1 | 1);
  b[rt] = b[rt << 1] + b[rt << 1 | 1];
}
int qry(int k, int l, int r, int rt) {
  if (l == r) return l;
  int m = l + r >> 1;
  if (b[rt << 1] >= k)
    return qry(k, l, m, rt << 1);
  else
    return qry(k - b[rt << 1], m + 1, r, rt << 1 | 1);
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); i++)
    scanf("%d", &a[i].first), a[i].second = i, c[i] = a[i].first;
  sort(a + 1, a + n + 1, [](pair<int, int> x, pair<int, int> y) {
    if (x.first != y.first)
      return x.first > y.first;
    else
      return x.second < y.second;
  });
  int m;
  scanf("%d", &m);
  for (int i = (1); i < (m + 1); i++) {
    scanf("%d%d", &q[i].k, &q[i].pos);
    q[i].id = i;
  }
  sort(q + 1, q + m + 1, cmp);
  int l = 0;
  for (int i = (1); i < (m + 1); i++) {
    while (l < q[i].k) l++, upd(a[l].second, 1, n, 1);
    ans[q[i].id] = c[qry(q[i].pos, 1, n, 1)];
  }
  for (int i = (1); i < (m + 1); i++) printf("%d\n", ans[i]);
  return 0;
}
