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 long long mod = 1000000007;
const long long inf = 1e18 * 4;
const long double pai = acos(-1);
int n;
int a[200009], b[200009];
int bef[200009];
vector<int> tree[800009];
vector<int> merge(vector<int> x, vector<int> y) {
  vector<int> ret = x;
  for (auto u : y) ret.push_back(u);
  sort(ret.begin(), ret.end());
  return ret;
}
void build(int node, int l, int r) {
  if (l == r) {
    tree[node].push_back(a[l]);
    return;
  }
  build(node * 2, l, (l + r) / 2);
  build(node * 2 + 1, (l + r) / 2 + 1, r);
  tree[node] = merge(tree[node * 2], tree[node * 2 + 1]);
}
int calc(int x, int node) {
  int l = -1, r = tree[node].size();
  while (r - l > 1) {
    if (tree[node][(l + r) / 2] >= x)
      r = (l + r) / 2;
    else
      l = (l + r) / 2;
  }
  return tree[node].size() - r;
}
int query(int node, int l, int r, int mn, int nummn, int crnt) {
  if (l == r) return a[l];
  int num = calc(mn + 1, node * 2);
  int X = calc(mn, node * 2);
  int xxx = min(X - num, nummn);
  num += xxx;
  if (num >= crnt)
    return query(node * 2, l, (l + r) / 2, mn, nummn, crnt);
  else
    return query(node * 2 + 1, (l + r) / 2 + 1, r, mn, nummn - xxx, crnt - num);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), b[i] = a[i];
  build(1, 0, n - 1);
  sort(b, b + n);
  reverse(b, b + n);
  int crnt = 0;
  for (int i = 0; i < n; i++) {
    int j = i, ret = 0;
    while (j < n && b[i] == b[j]) {
      bef[j] = crnt;
      ret++;
      j++;
    }
    crnt += ret;
    i = j - 1;
  }
  int q;
  cin >> q;
  while (q--) {
    int x, k;
    scanf("%d", &x);
    scanf("%d", &k);
    int mn = b[x - 1];
    int nummn = x - bef[x - 1];
    printf("%d\n", query(1, 0, n - 1, mn, nummn, k));
  }
}
