#include <bits/stdc++.h>
const double pi = acos(-1.0);
const int maxn = (2e5 + 10);
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
using namespace std;
struct node {
  int val;
  int id;
} a[maxn];
bool cmp(node a, node b) {
  if (a.val == b.val) {
    return a.id < b.id;
  }
  return a.val > b.val;
}
struct nod {
  int sum;
  int pos;
  int id;
} q[maxn];
bool cmpp(nod a, nod b) { return a.sum < b.sum; }
int ans[maxn];
int tree[maxn * 4];
void push_up(int node) { tree[node] = tree[node << 1] + tree[node << 1 | 1]; }
void update(int node, int start, int end, int val) {
  if (start == end) {
    tree[node] += 1;
    return;
  }
  int mid = (start + end) / 2;
  if (val <= mid) update(node * 2, start, mid, val);
  if (val > mid) update(node * 2 + 1, mid + 1, end, val);
  push_up(node);
}
int query(int node, int start, int end, int k) {
  if (start == end) return start;
  int mid = (start + end) / 2;
  if (tree[node << 1] >= k)
    return query(node * 2, start, mid, k);
  else
    return query(node * 2 + 1, mid + 1, end, k - tree[node << 1]);
}
map<int, int> ys;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    node now;
    now.id = i;
    now.val = x;
    ys[i] = x;
    a[i] = now;
  }
  sort(a + 1, a + 1 + n, cmp);
  int s;
  cin >> s;
  for (int i = 1; i <= s; i++) {
    int x, y;
    cin >> x >> y;
    nod now;
    now.sum = x;
    now.pos = y;
    now.id = i;
    q[i] = now;
  }
  sort(q + 1, q + 1 + s, cmpp);
  int xz = 0;
  for (int i = 1; i <= s; i++) {
    while (xz < q[i].sum) {
      xz += 1;
      update(1, 1, n, a[xz].id);
    }
    int now = query(1, 1, n, q[i].pos);
    ans[q[i].id] = ys[now];
  }
  for (int i = 1; i <= s; i++) cout << ans[i] << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
const double pi = acos(-1.0);
const int maxn = (2e5 + 10);
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
using namespace std;
struct node {
  int val;
  int id;
} a[maxn];
bool cmp(node a, node b) {
  if (a.val == b.val) {
    return a.id < b.id;
  }
  return a.val > b.val;
}
struct nod {
  int sum;
  int pos;
  int id;
} q[maxn];
bool cmpp(nod a, nod b) { return a.sum < b.sum; }
int ans[maxn];
int tree[maxn * 4];
void push_up(int node) { tree[node] = tree[node << 1] + tree[node << 1 | 1]; }
void update(int node, int start, int end, int val) {
  if (start == end) {
    tree[node] += 1;
    return;
  }
  int mid = (start + end) / 2;
  if (val <= mid) update(node * 2, start, mid, val);
  if (val > mid) update(node * 2 + 1, mid + 1, end, val);
  push_up(node);
}
int query(int node, int start, int end, int k) {
  if (start == end) return start;
  int mid = (start + end) / 2;
  if (tree[node << 1] >= k)
    return query(node * 2, start, mid, k);
  else
    return query(node * 2 + 1, mid + 1, end, k - tree[node << 1]);
}
map<int, int> ys;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    node now;
    now.id = i;
    now.val = x;
    ys[i] = x;
    a[i] = now;
  }
  sort(a + 1, a + 1 + n, cmp);
  int s;
  cin >> s;
  for (int i = 1; i <= s; i++) {
    int x, y;
    cin >> x >> y;
    nod now;
    now.sum = x;
    now.pos = y;
    now.id = i;
    q[i] = now;
  }
  sort(q + 1, q + 1 + s, cmpp);
  int xz = 0;
  for (int i = 1; i <= s; i++) {
    while (xz < q[i].sum) {
      xz += 1;
      update(1, 1, n, a[xz].id);
    }
    int now = query(1, 1, n, q[i].pos);
    ans[q[i].id] = ys[now];
  }
  for (int i = 1; i <= s; i++) cout << ans[i] << endl;
  return 0;
}

```