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;
struct vec {
  long double x, y;
  bool operator==(const vec &a) const { return x == a.x && y == a.y; }
  bool operator<(const vec &a) const {
    if (a.x != x) return x < a.x;
    return y < a.y;
  }
  vec operator+(const vec &a) const { return {x + a.x, y + a.y}; }
  vec operator*(const long double &a) const { return {x * a, y * a}; }
  vec operator-(const vec &a) const { return {x - a.x, y - a.y}; }
  vec operator/(const long double &a) const { return {x / a, y / a}; }
  long double len() const { return (x * x + y * y); }
  long double dot(const vec &a) const { return x * a.x + y * a.y; }
  long double crs(const vec &a) const { return x * a.y - y * a.x; }
  vec proj(const vec &a) const {
    return (a / a.len()) * (*this).dot(a) / a.len();
  }
};
struct P {
  int x, y, z;
  bool operator<(const P &a) const { return x > a.x; }
};
bool as(P a, P b) {
  if (a.x != b.x) return a.x > b.x;
  return a.y < b.y;
}
long long x, y, z, mod = 1000000007;
vector<int> v[2];
int i, n, m, k, a, d, b, c, h, dx[10] = {1, 0, -1, 0}, dy[10] = {0, 1, 0, -1};
int e;
int o[511221];
int l[511111];
int j[211111];
stack<int> s;
queue<P> q, q1;
P u[200101], u1[201100];
char r[2221];
int g(int n) {
  int c = 1;
  for (; c < i;) {
    if (l[c * 2] >= n)
      c *= 2;
    else {
      n -= l[c * 2];
      c = c * 2 + 1;
    }
  }
  return c - i + 1;
}
int main() {
  scanf("%d", &a);
  for (int t = 1; t <= a; t++) {
    scanf("%d", &o[t]);
    u[t] = {o[t], t};
  }
  for (i = 1; i < a; i *= 2)
    ;
  scanf("%d", &b);
  for (int t = 1; t <= b; t++) {
    scanf("%d %d", &n, &m);
    u1[t] = {n, m, t};
  }
  sort(u + 1, u + a + 1, as);
  sort(u1 + 1, u1 + b + 1, as);
  e = 1;
  for (int t = b; t >= 1; t--) {
    for (; e <= a && e <= u1[t].x; e++) {
      l[i + u[e].y - 1]++;
      for (int h = (i + u[e].y - 1) / 2; h; h /= 2) l[h]++;
    }
    j[u1[t].z] = o[g(u1[t].y)];
  }
  for (int t = 1; t <= b; t++) printf("%d\n", j[t]);
}
