#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 51 + 51, MOD = 1e9 + 7, mod = 998244353, INF = 2e18;
mt19937 Flash_new_51(51515);
using vi = vector<long long>;
using vp = vector<pair<long long, long long>>;
struct Node {
  long long k, pos, ind;
  Node() { k = pos = ind = 0; }
  Node(long long k1, long long pos1, long long ind1) {
    k = k1, pos = pos1, ind = ind1;
  }
};
vector<Node> kek;
bool cmp(Node A, Node B) {
  if (A.k == B.k) {
    return A.pos < B.pos;
  }
  return A.k < B.k;
}
long long t[N], n, ans[N], b[N];
pair<long long, long long> a[N];
bool used[N];
void inc(long long i, long long delta) {
  used[i] = 1;
  for (; i < n; i |= (i + 1)) {
    t[i] += delta;
  }
}
long long sum(long long i) {
  long long summ = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) summ += t[i];
  return summ;
}
long long zap(long long pos) {
  long long l = 0, r = n;
  while (r - l > 1) {
    long long m = (r + l) >> 1;
    long long k = sum(m);
    if (k == pos) {
      if (used[m])
        return b[m];
      else
        r = m;
    } else {
      if (k < pos)
        l = m;
      else
        r = m;
    }
  }
  return b[l];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = -i;
    b[i] = a[i].first;
  }
  sort(a, a + n);
  reverse(a, a + n);
  long long q;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long k, pos;
    cin >> k >> pos;
    kek.push_back({k, pos, i});
  }
  sort(kek.begin(), kek.end(), cmp);
  long long tek = 0;
  for (auto to : kek) {
    for (long long i = tek; i < to.k; i++) {
      inc(abs(a[i].second), 1);
    }
    tek = to.k;
    ans[to.ind] = zap(to.pos);
  }
  for (long long i = 0; i < q; i++) cout << ans[i] << "\n";
}

### 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 long long N = 1e6 + 51 + 51, MOD = 1e9 + 7, mod = 998244353, INF = 2e18;
mt19937 Flash_new_51(51515);
using vi = vector<long long>;
using vp = vector<pair<long long, long long>>;
struct Node {
  long long k, pos, ind;
  Node() { k = pos = ind = 0; }
  Node(long long k1, long long pos1, long long ind1) {
    k = k1, pos = pos1, ind = ind1;
  }
};
vector<Node> kek;
bool cmp(Node A, Node B) {
  if (A.k == B.k) {
    return A.pos < B.pos;
  }
  return A.k < B.k;
}
long long t[N], n, ans[N], b[N];
pair<long long, long long> a[N];
bool used[N];
void inc(long long i, long long delta) {
  used[i] = 1;
  for (; i < n; i |= (i + 1)) {
    t[i] += delta;
  }
}
long long sum(long long i) {
  long long summ = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) summ += t[i];
  return summ;
}
long long zap(long long pos) {
  long long l = 0, r = n;
  while (r - l > 1) {
    long long m = (r + l) >> 1;
    long long k = sum(m);
    if (k == pos) {
      if (used[m])
        return b[m];
      else
        r = m;
    } else {
      if (k < pos)
        l = m;
      else
        r = m;
    }
  }
  return b[l];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = -i;
    b[i] = a[i].first;
  }
  sort(a, a + n);
  reverse(a, a + n);
  long long q;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long k, pos;
    cin >> k >> pos;
    kek.push_back({k, pos, i});
  }
  sort(kek.begin(), kek.end(), cmp);
  long long tek = 0;
  for (auto to : kek) {
    for (long long i = tek; i < to.k; i++) {
      inc(abs(a[i].second), 1);
    }
    tek = to.k;
    ans[to.ind] = zap(to.pos);
  }
  for (long long i = 0; i < q; i++) cout << ans[i] << "\n";
}

```