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;
namespace IO {
const int BS = (1 << 23) + 5;
int Top = 0;
char Buffer[BS], OT[BS], *OS = OT, *HD, *TL, SS[20];
const char *fin = OT + BS - 1;
char Getchar() {
  if (HD == TL) {
    TL = (HD = Buffer) + fread(Buffer, 1, BS, stdin);
  }
  return (HD == TL) ? EOF : *HD++;
}
void flush() { fwrite(OT, 1, OS - OT, stdout), OS = OT; }
void Putchar(char c) {
  *OS++ = c;
  if (OS == fin) flush();
}
void write(long long x) {
  if (!x) {
    Putchar('0');
    return;
  }
  if (x < 0) x = -x, Putchar('-');
  while (x) SS[++Top] = x % 10, x /= 10;
  while (Top) Putchar(SS[Top] + '0'), --Top;
}
inline long long read() {
  long long nm = 0;
  bool fh = true;
  char cw = getchar();
  for (; !isdigit(cw); cw = getchar()) fh ^= (cw == '-');
  for (; isdigit(cw); cw = getchar()) nm = nm * 10 + (cw - '0');
  return fh ? nm : -nm;
}
}  // namespace IO
using namespace IO;
int n, m;
int p[200200], cnt;
int L[200200 * 30], R[200200 * 30], sm[200200 * 30], rt[200200];
pair<int, int> t[200200];
inline void ins(int &x, int y, int l, int r, int pos) {
  x = ++cnt, sm[x] = sm[y] + 1, L[x] = L[y], R[x] = R[y];
  if (l == r) return;
  int mid = ((l + r) >> 1);
  (pos <= mid) ? ins(L[x], L[y], l, mid, pos)
               : ins(R[x], R[y], mid + 1, r, pos);
}
inline int fd(int x, int l, int r, int rk) {
  if (l == r) return p[l];
  int mid = ((l + r) >> 1);
  if (sm[L[x]] >= rk) return fd(L[x], l, mid, rk);
  return fd(R[x], mid + 1, r, rk - sm[L[x]]);
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) t[i] = make_pair(-(p[i] = read()), i);
  sort(t + 1, t + n + 1);
  for (int i = 1; i <= n; i++) ins(rt[i], rt[i - 1], 1, n, t[i].second);
  for (int k, w, Cas = read(); Cas; --Cas) {
    k = read(), w = read();
    printf("%d\n", fd(rt[k], 1, n, w));
  }
  return 0;
}
