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 int MOD = 1e9 + 7;
long long REM(long long a) {
  a %= MOD;
  if (a < 0) a += MOD;
  return a;
}
long long ADD(long long a, long long b) { return REM(REM(a) + REM(b)); }
long long SUB(long long a, long long b) { return REM(REM(a) - REM(b)); }
long long MUL(long long a, long long b) { return REM(REM(a) * REM(b)); }
long long POT(long long n, long long k) {
  if (k == 0) return 1;
  long long ret = POT(n, k / 2);
  ret = MUL(ret, ret);
  if (k % 2) return MUL(ret, n);
  return ret;
}
long long DIV(long long a, long long b) { return MUL(a, POT(b, MOD - 2)); }
bool isPrime(long long x) {
  for (long long i = 2; i <= sqrt(x); i++)
    if (x % i == 0) return false;
  return true;
}
const int MAXN = 2e5 + 10;
int n, m;
int A[MAXN];
vector<pair<int, int> > prec;
vector<int> prec1;
vector<pair<int, int> > q[MAXN];
int sol[MAXN];
int fwt[MAXN];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.second != b.second) return a.second < b.second;
  return a.first < b.first;
}
void update(int x, int v) {
  for (; x < MAXN; x += (x & -x)) fwt[x] += v;
}
int query(int x) {
  int ret = 0;
  for (; x > 0; x -= (x & -x)) ret += fwt[x];
  return ret;
}
int binary(int x) {
  int lo = 1, hi = n, ans = -1;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    int s = query(mid);
    if (s >= x) {
      ans = mid;
      hi = mid - 1;
    } else
      lo = mid + 1;
  }
  return ans;
}
int main() {
  cin >> n;
  multiset<pair<int, int> > ms;
  for (int i = (0); i < (n); i++) {
    cin >> A[i];
    ms.insert(make_pair(A[i], -i));
    update(i + 1, 1);
  }
  cin >> m;
  for (int i = (0); i < (m); i++) {
    int k, pos;
    cin >> k >> pos;
    q[k].push_back(make_pair(pos, i));
  }
  for (int i = (n); i >= (1); i--) {
    pair<int, int> x = (*ms.begin());
    ms.erase(ms.find(x));
    x.second *= -1;
    for (int j = (0); j < ((int)q[i].size()); j++) {
      int pos = q[i][j].first;
      int pl = q[i][j].second;
      int pos1 = binary(pos);
      sol[pl] = A[pos1 - 1];
    }
    update(x.second + 1, -1);
  }
  for (int i = (0); i < (m); i++) cout << sol[i] << "\n";
  return 0;
}
