#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 10;
int n, m, a[N], ans[N];
vector<pair<int, int> > g[N];
pair<int, int> b[N];
int tr[N << 2];
void build(int o, int l, int r) {
  tr[o] = r - l + 1;
  if (l != r)
    build((o << 1), l, ((l + r) >> 1)),
        build(((o << 1) | 1), ((l + r) >> 1) + 1, r);
}
void upd(int o, int l, int r, int first) {
  --tr[o];
  if (l != r)
    ((l + r) >> 1) >= first ? upd((o << 1), l, ((l + r) >> 1), first)
                            : upd(((o << 1) | 1), ((l + r) >> 1) + 1, r, first);
}
int find(int o, int l, int r, int k) {
  if (l == r) return a[l];
  if (k > tr[(o << 1)])
    return find(((o << 1) | 1), ((l + r) >> 1) + 1, r, k - tr[(o << 1)]);
  return find((o << 1), l, ((l + r) >> 1), k);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%d", a + i), b[i].first = a[i], b[i].second = -i;
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int k, pos;
    scanf("%d%d", &k, &pos);
    g[k].push_back({pos, i});
  }
  build(1, 1, n);
  sort(b + 1, b + 1 + n);
  for (int i = 1; i <= n; ++i) {
    for (auto &t : g[n - i + 1]) ans[t.second] = find(1, 1, n, t.first);
    upd(1, 1, n, -b[i].second);
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
}

### Prompt

Your task is to create a Cpp solution 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;
const int P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 10;
int n, m, a[N], ans[N];
vector<pair<int, int> > g[N];
pair<int, int> b[N];
int tr[N << 2];
void build(int o, int l, int r) {
  tr[o] = r - l + 1;
  if (l != r)
    build((o << 1), l, ((l + r) >> 1)),
        build(((o << 1) | 1), ((l + r) >> 1) + 1, r);
}
void upd(int o, int l, int r, int first) {
  --tr[o];
  if (l != r)
    ((l + r) >> 1) >= first ? upd((o << 1), l, ((l + r) >> 1), first)
                            : upd(((o << 1) | 1), ((l + r) >> 1) + 1, r, first);
}
int find(int o, int l, int r, int k) {
  if (l == r) return a[l];
  if (k > tr[(o << 1)])
    return find(((o << 1) | 1), ((l + r) >> 1) + 1, r, k - tr[(o << 1)]);
  return find((o << 1), l, ((l + r) >> 1), k);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%d", a + i), b[i].first = a[i], b[i].second = -i;
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int k, pos;
    scanf("%d%d", &k, &pos);
    g[k].push_back({pos, i});
  }
  build(1, 1, n);
  sort(b + 1, b + 1 + n);
  for (int i = 1; i <= n; ++i) {
    for (auto &t : g[n - i + 1]) ans[t.second] = find(1, 1, n, t.first);
    upd(1, 1, n, -b[i].second);
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
}

```