#include <bits/stdc++.h>
using namespace std;
void solve() {}
const long long maxn = 200500;
long long a[maxn];
long long seg[3 * maxn];
long long n;
void upd(long long pos, long long val) {
  pos += 2e5;
  seg[pos] = val;
  for (long long i = pos / 2; i >= 1; i /= 2) {
    seg[i] = seg[i * 2] + seg[i * 2 + 1];
  }
}
long long query(long long l, long long r) {
  l += 2e5, r += 2e5;
  long long res = 0;
  while (l <= r) {
    if (l % 2 == 1) res += seg[l++];
    if (r % 2 == 0) res += seg[r--];
    l /= 2;
    r /= 2;
  }
  return res;
}
set<long long> S;
map<long long, long long> cmp, rcmp;
long long curc = 1;
bool comp(pair<long long, long long> &l, pair<long long, long long> &r) {
  if (l.first == r.first) {
    return l.second < r.second;
  }
  return l.first > r.first;
}
struct Query {
  long long k, pos, ind;
  Query(long long kk, long long poss, long long indd) {
    k = kk;
    pos = poss;
    ind = indd;
  }
};
bool compQuery(Query &l, Query &r) { return l.k < r.k; }
vector<Query> queries;
long long ans[maxn];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long> > vals;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    vals.emplace_back(a[i], i);
    S.emplace(a[i]);
  }
  for (long long x : S) {
    cmp[x] = curc;
    rcmp[curc] = x;
    curc++;
  }
  for (long long i = 0; i < (long long)vals.size(); i++) {
    vals[i].first = cmp[vals[i].first];
  }
  sort((vals).begin(), (vals).end(), comp);
  for (pair<long long, long long> x : vals) {
    if (false) cerr << x.first << " " << x.second << '\n';
  }
  long long m;
  cin >> m;
  for (long long i = 1; i <= m; i++) {
    long long k, pos;
    cin >> k >> pos;
    queries.emplace_back(k, pos, i);
  }
  sort((queries).begin(), (queries).end(), compQuery);
  long long cur = 0;
  for (Query q : queries) {
    if (false)
      cerr << "processing query " << q.k << " " << q.pos << " " << q.ind
           << '\n';
    while (cur < q.k) {
      if (false) cerr << "updating to " << vals[cur].second << '\n';
      long long pre = query(vals[cur].second, vals[cur].second);
      upd(vals[cur].second, pre + 1);
      if (false) cerr << ("query(1, 2)") << " is " << (query(1, 2)) << '\n';
      cur++;
    }
    long long l = 1;
    long long r = 2e5;
    long long lgood = -1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (query(1, mid) >= q.pos) {
        lgood = mid;
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    assert(lgood != -1);
    if (false) cerr << ("lgood") << " is " << (lgood) << '\n';
    if (false)
      cerr << ("query(1, lgood)") << " is " << (query(1, lgood)) << '\n';
    if (false) cerr << ("query(1, 2)") << " is " << (query(1, 2)) << '\n';
    if (false) cerr << "ans is " << a[lgood] << '\n';
    ans[q.ind] = a[lgood];
    if (false) cerr << '\n';
  }
  for (long long i = 1; i <= m; i++) {
    cout << ans[i] << '\n';
  }
}

### Prompt

Please formulate 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;
void solve() {}
const long long maxn = 200500;
long long a[maxn];
long long seg[3 * maxn];
long long n;
void upd(long long pos, long long val) {
  pos += 2e5;
  seg[pos] = val;
  for (long long i = pos / 2; i >= 1; i /= 2) {
    seg[i] = seg[i * 2] + seg[i * 2 + 1];
  }
}
long long query(long long l, long long r) {
  l += 2e5, r += 2e5;
  long long res = 0;
  while (l <= r) {
    if (l % 2 == 1) res += seg[l++];
    if (r % 2 == 0) res += seg[r--];
    l /= 2;
    r /= 2;
  }
  return res;
}
set<long long> S;
map<long long, long long> cmp, rcmp;
long long curc = 1;
bool comp(pair<long long, long long> &l, pair<long long, long long> &r) {
  if (l.first == r.first) {
    return l.second < r.second;
  }
  return l.first > r.first;
}
struct Query {
  long long k, pos, ind;
  Query(long long kk, long long poss, long long indd) {
    k = kk;
    pos = poss;
    ind = indd;
  }
};
bool compQuery(Query &l, Query &r) { return l.k < r.k; }
vector<Query> queries;
long long ans[maxn];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long> > vals;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    vals.emplace_back(a[i], i);
    S.emplace(a[i]);
  }
  for (long long x : S) {
    cmp[x] = curc;
    rcmp[curc] = x;
    curc++;
  }
  for (long long i = 0; i < (long long)vals.size(); i++) {
    vals[i].first = cmp[vals[i].first];
  }
  sort((vals).begin(), (vals).end(), comp);
  for (pair<long long, long long> x : vals) {
    if (false) cerr << x.first << " " << x.second << '\n';
  }
  long long m;
  cin >> m;
  for (long long i = 1; i <= m; i++) {
    long long k, pos;
    cin >> k >> pos;
    queries.emplace_back(k, pos, i);
  }
  sort((queries).begin(), (queries).end(), compQuery);
  long long cur = 0;
  for (Query q : queries) {
    if (false)
      cerr << "processing query " << q.k << " " << q.pos << " " << q.ind
           << '\n';
    while (cur < q.k) {
      if (false) cerr << "updating to " << vals[cur].second << '\n';
      long long pre = query(vals[cur].second, vals[cur].second);
      upd(vals[cur].second, pre + 1);
      if (false) cerr << ("query(1, 2)") << " is " << (query(1, 2)) << '\n';
      cur++;
    }
    long long l = 1;
    long long r = 2e5;
    long long lgood = -1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (query(1, mid) >= q.pos) {
        lgood = mid;
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    assert(lgood != -1);
    if (false) cerr << ("lgood") << " is " << (lgood) << '\n';
    if (false)
      cerr << ("query(1, lgood)") << " is " << (query(1, lgood)) << '\n';
    if (false) cerr << ("query(1, 2)") << " is " << (query(1, 2)) << '\n';
    if (false) cerr << "ans is " << a[lgood] << '\n';
    ans[q.ind] = a[lgood];
    if (false) cerr << '\n';
  }
  for (long long i = 1; i <= m; i++) {
    cout << ans[i] << '\n';
  }
}

```