#include <bits/stdc++.h>
using namespace std;
const int N = (int)2e5 + 2;
const int LG = 20;
const int mod = (int)1e9 + 7;
const int INF = (int)1e9 + 7;
const long long linf = 1e16;
const long double pi = acos(-1);
const int base = 10;
const long double EPS = 1e-9;
int n, m, a[N], f[N], ans[N];
pair<int, pair<int, int> > q[N];
set<pair<int, int> > st;
void update(int pos, int val) {
  for (; pos <= N; pos |= pos + 1) f[pos] += val;
}
int get(int pos) {
  int sum = 0;
  for (; pos >= 0; pos &= pos + 1, pos--) sum += f[pos];
  return sum;
}
int calc(int pos) {
  int l = 0, r = N;
  while (r - l > 1) {
    int mid = r + l >> 1;
    if (get(mid) >= pos)
      r = mid;
    else
      l = mid;
  }
  if (get(r) >= pos)
    return r;
  else
    return l;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    st.insert({-a[i], i});
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d %d", &q[i].first, &q[i].second.first);
    q[i].second.second = i;
  }
  sort(q + 1, q + m + 1);
  vector<int> v;
  int cursum = 0;
  for (int i = 1; i <= m; ++i) {
    if (v.size() != q[i].first) {
      while (v.size() < q[i].first) {
        int to = st.begin()->second;
        update(to, 1);
        v.push_back(to);
        cursum += (abs(st.begin()->first));
        st.erase(st.begin());
      }
    }
    int res = calc(q[i].second.first);
    ans[q[i].second.second] = a[res];
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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]. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = (int)2e5 + 2;
const int LG = 20;
const int mod = (int)1e9 + 7;
const int INF = (int)1e9 + 7;
const long long linf = 1e16;
const long double pi = acos(-1);
const int base = 10;
const long double EPS = 1e-9;
int n, m, a[N], f[N], ans[N];
pair<int, pair<int, int> > q[N];
set<pair<int, int> > st;
void update(int pos, int val) {
  for (; pos <= N; pos |= pos + 1) f[pos] += val;
}
int get(int pos) {
  int sum = 0;
  for (; pos >= 0; pos &= pos + 1, pos--) sum += f[pos];
  return sum;
}
int calc(int pos) {
  int l = 0, r = N;
  while (r - l > 1) {
    int mid = r + l >> 1;
    if (get(mid) >= pos)
      r = mid;
    else
      l = mid;
  }
  if (get(r) >= pos)
    return r;
  else
    return l;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    st.insert({-a[i], i});
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d %d", &q[i].first, &q[i].second.first);
    q[i].second.second = i;
  }
  sort(q + 1, q + m + 1);
  vector<int> v;
  int cursum = 0;
  for (int i = 1; i <= m; ++i) {
    if (v.size() != q[i].first) {
      while (v.size() < q[i].first) {
        int to = st.begin()->second;
        update(to, 1);
        v.push_back(to);
        cursum += (abs(st.begin()->first));
        st.erase(st.begin());
      }
    }
    int res = calc(q[i].second.first);
    ans[q[i].second.second] = a[res];
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}

```