#include <bits/stdc++.h>
using namespace std;
template <typename F, typename T>
struct lazy_segment_tree {
  F merge;
  T identity;
  vector<T> data, lazy;
  int n;
  lazy_segment_tree(int len, F _merge, T _identity)
      : merge(_merge), identity(_identity) {
    n = pow(2, ceil(log2(len)));
    data.resize(2 * n - 1, identity);
    lazy.resize(2 * n - 1, 0);
  }
  void init(int k, T p) {
    k += n - 1;
    data[k] = p;
    while (k > 0) {
      k = (k - 1) / 2;
      data[k] = merge(data[k * 2 + 1], data[k * 2 + 2]);
    }
  }
  void add(int l, int r, int x) { return add(l, r, x, 0, 0, n); }
  void add(int a, int b, int x, int k, int l, int r) {
    if (r <= a || b <= l) return;
    if (a <= l && r <= b) {
      lazy[k] += x;
      return;
    }
    int m = (l + r) / 2;
    add(a, b, x, k * 2 + 1, l, m);
    add(a, b, x, k * 2 + 2, m, r);
    data[k] = merge(data[k * 2 + 1] + lazy[k * 2 + 1],
                    data[k * 2 + 2] + lazy[k * 2 + 2]);
  }
  T query(int l, int r) { return query(l, r, 0, 0, n); }
  T query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return identity;
    if (a <= l && r <= b) return data[k] + lazy[k];
    int m = (l + r) / 2;
    T left = query(a, b, k * 2 + 1, l, m);
    T right = query(a, b, k * 2 + 2, m, r);
    return merge(left, right) + lazy[k];
  }
  int find(T val) { return unko(val, 0, 0, n); }
  int unko(T val, int k, int l, int r) {
    if (r - l > 1) {
      int m = (l + r) / 2;
      T left = query(l, m);
      if (val <= left) {
        return unko(val, k * 2 + 1, l, m);
      } else {
        return unko(val, k * 2 + 2, m, r);
      }
    } else {
      T v = query(l, l + 1);
      if (v == val) {
        return l;
      } else {
        return -1;
      }
    }
  }
};
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; (i) < (int)(n); (i)++) {
    scanf("%d", &a[i]);
  }
  int m;
  scanf("%d", &m);
  vector<vector<pair<int, int>>> query(n);
  for (int i = 0; (i) < (int)(m); (i)++) {
    int k, pos;
    scanf("%d%d", &k, &pos);
    pos--;
    k--;
    query[k].emplace_back(pos, i);
  }
  map<int, int> unko;
  vector<int> nxt(n, -1);
  for (int i = n - 1; i >= 0; i--) {
    if (unko.count(a[i])) nxt[i] = unko[a[i]];
    unko[a[i]] = i;
  }
  vector<int> benki;
  for (auto it : unko) benki.push_back(it.first);
  sort(benki.rbegin(), benki.rend());
  auto f = [&](int a, int b) { return max(a, b); };
  lazy_segment_tree<decltype(f), int> segtree(n, f, -1e8);
  int p = 0;
  vector<int> ans(m);
  for (int b : benki) {
    int pos = unko[b];
    while (true) {
      int idx = segtree.query(0, pos);
      idx++;
      if (idx < 0) idx = 0;
      segtree.add(pos, pos + 1, -segtree.query(pos, pos + 1) + idx);
      segtree.add(pos + 1, n, 1);
      pos = nxt[pos];
      for (auto q : query[p]) {
        int w = segtree.find(q.first);
        ans[q.second] = a[w];
      }
      p++;
      if (pos == -1) {
        break;
      }
    }
  }
  assert(p == n);
  for (int i = 0; (i) < (int)(m); (i)++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Please create a solution in CPP to 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;
template <typename F, typename T>
struct lazy_segment_tree {
  F merge;
  T identity;
  vector<T> data, lazy;
  int n;
  lazy_segment_tree(int len, F _merge, T _identity)
      : merge(_merge), identity(_identity) {
    n = pow(2, ceil(log2(len)));
    data.resize(2 * n - 1, identity);
    lazy.resize(2 * n - 1, 0);
  }
  void init(int k, T p) {
    k += n - 1;
    data[k] = p;
    while (k > 0) {
      k = (k - 1) / 2;
      data[k] = merge(data[k * 2 + 1], data[k * 2 + 2]);
    }
  }
  void add(int l, int r, int x) { return add(l, r, x, 0, 0, n); }
  void add(int a, int b, int x, int k, int l, int r) {
    if (r <= a || b <= l) return;
    if (a <= l && r <= b) {
      lazy[k] += x;
      return;
    }
    int m = (l + r) / 2;
    add(a, b, x, k * 2 + 1, l, m);
    add(a, b, x, k * 2 + 2, m, r);
    data[k] = merge(data[k * 2 + 1] + lazy[k * 2 + 1],
                    data[k * 2 + 2] + lazy[k * 2 + 2]);
  }
  T query(int l, int r) { return query(l, r, 0, 0, n); }
  T query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return identity;
    if (a <= l && r <= b) return data[k] + lazy[k];
    int m = (l + r) / 2;
    T left = query(a, b, k * 2 + 1, l, m);
    T right = query(a, b, k * 2 + 2, m, r);
    return merge(left, right) + lazy[k];
  }
  int find(T val) { return unko(val, 0, 0, n); }
  int unko(T val, int k, int l, int r) {
    if (r - l > 1) {
      int m = (l + r) / 2;
      T left = query(l, m);
      if (val <= left) {
        return unko(val, k * 2 + 1, l, m);
      } else {
        return unko(val, k * 2 + 2, m, r);
      }
    } else {
      T v = query(l, l + 1);
      if (v == val) {
        return l;
      } else {
        return -1;
      }
    }
  }
};
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; (i) < (int)(n); (i)++) {
    scanf("%d", &a[i]);
  }
  int m;
  scanf("%d", &m);
  vector<vector<pair<int, int>>> query(n);
  for (int i = 0; (i) < (int)(m); (i)++) {
    int k, pos;
    scanf("%d%d", &k, &pos);
    pos--;
    k--;
    query[k].emplace_back(pos, i);
  }
  map<int, int> unko;
  vector<int> nxt(n, -1);
  for (int i = n - 1; i >= 0; i--) {
    if (unko.count(a[i])) nxt[i] = unko[a[i]];
    unko[a[i]] = i;
  }
  vector<int> benki;
  for (auto it : unko) benki.push_back(it.first);
  sort(benki.rbegin(), benki.rend());
  auto f = [&](int a, int b) { return max(a, b); };
  lazy_segment_tree<decltype(f), int> segtree(n, f, -1e8);
  int p = 0;
  vector<int> ans(m);
  for (int b : benki) {
    int pos = unko[b];
    while (true) {
      int idx = segtree.query(0, pos);
      idx++;
      if (idx < 0) idx = 0;
      segtree.add(pos, pos + 1, -segtree.query(pos, pos + 1) + idx);
      segtree.add(pos + 1, n, 1);
      pos = nxt[pos];
      for (auto q : query[p]) {
        int w = segtree.find(q.first);
        ans[q.second] = a[w];
      }
      p++;
      if (pos == -1) {
        break;
      }
    }
  }
  assert(p == n);
  for (int i = 0; (i) < (int)(m); (i)++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```