#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long double eps = 1e-12;
long long dx[] = {0, 0, 1, -1};
long long dy[] = {1, -1, 0, 0};
long long n, res = 0, k = 0, h, ans = 0, m, level = 0;
;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
void yes() { cout << "Yes\n"; }
void no() { cout << "No\n"; }
vector<long long> v[300005], vnode;
void add_edge(long long x, long long y) {
  v[x].push_back(y);
  v[y].push_back(x);
}
string s2, ch = "", s = "", s1 = "", t;
vector<pair<string, long long> > vm;
vector<pair<pair<long long, long long>, long long> > vvvl, vvvr;
vector<pair<long long, long long> > vv1, vv;
long long a[500005], b[500005], dp[500004];
map<long long, long long> mp, mprev;
long long bit[1000005] = {0};
void update(long long x, long long val) {
  for (; x <= 1e6; x += x & -x) bit[x] += val;
}
long long query(long long x) {
  long long sum = 0;
  for (; x > 0; x -= x & -x) {
    sum += bit[x];
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long i = 0, x = 0, z = 0, y = 0, j = 0, q, mx = 0, mm, idx = 0, ok = 0,
            l = 0, r, negatif = 0, positif = 0, l1, r1, p;
  set<long long>::iterator it, it2;
  multiset<long long>::iterator it1;
  multiset<long long> mst1, mst2;
  cin >> n;
  set<long long> st;
  for (i = 1; i <= (long long)(n); i++) {
    cin >> a[i];
    vv.push_back({a[i], -i});
  }
  sort(vv.begin(), vv.end());
  cin >> m;
  for (i = 1; i <= m; i++) {
    cin >> x >> b[i];
    vv1.push_back({x, i});
  }
  sort(vv1.begin(), vv1.end());
  long long cnt = 0;
  j = n - 1;
  for (i = 0; i < vv1.size(); i++) {
    long long nb = vv1[i].first;
    long long idx = vv1[i].second;
    while (cnt < nb && j >= 0) {
      long long idx2 = -vv[j].second;
      j--;
      update(idx2, 1);
      cnt++;
    }
    long long pos = b[idx];
    l = 1;
    r = n;
    ans = 0;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (query(mid) < pos) {
        l = mid + 1;
        ans = mid;
      } else {
        r = mid - 1;
      }
    }
    ans++;
    dp[idx] = ans;
  }
  for (i = 1; i <= (long long)(m); i++) cout << a[dp[i]] << '\n';
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
long long mod = 1e9 + 7;
long double eps = 1e-12;
long long dx[] = {0, 0, 1, -1};
long long dy[] = {1, -1, 0, 0};
long long n, res = 0, k = 0, h, ans = 0, m, level = 0;
;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
void yes() { cout << "Yes\n"; }
void no() { cout << "No\n"; }
vector<long long> v[300005], vnode;
void add_edge(long long x, long long y) {
  v[x].push_back(y);
  v[y].push_back(x);
}
string s2, ch = "", s = "", s1 = "", t;
vector<pair<string, long long> > vm;
vector<pair<pair<long long, long long>, long long> > vvvl, vvvr;
vector<pair<long long, long long> > vv1, vv;
long long a[500005], b[500005], dp[500004];
map<long long, long long> mp, mprev;
long long bit[1000005] = {0};
void update(long long x, long long val) {
  for (; x <= 1e6; x += x & -x) bit[x] += val;
}
long long query(long long x) {
  long long sum = 0;
  for (; x > 0; x -= x & -x) {
    sum += bit[x];
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long i = 0, x = 0, z = 0, y = 0, j = 0, q, mx = 0, mm, idx = 0, ok = 0,
            l = 0, r, negatif = 0, positif = 0, l1, r1, p;
  set<long long>::iterator it, it2;
  multiset<long long>::iterator it1;
  multiset<long long> mst1, mst2;
  cin >> n;
  set<long long> st;
  for (i = 1; i <= (long long)(n); i++) {
    cin >> a[i];
    vv.push_back({a[i], -i});
  }
  sort(vv.begin(), vv.end());
  cin >> m;
  for (i = 1; i <= m; i++) {
    cin >> x >> b[i];
    vv1.push_back({x, i});
  }
  sort(vv1.begin(), vv1.end());
  long long cnt = 0;
  j = n - 1;
  for (i = 0; i < vv1.size(); i++) {
    long long nb = vv1[i].first;
    long long idx = vv1[i].second;
    while (cnt < nb && j >= 0) {
      long long idx2 = -vv[j].second;
      j--;
      update(idx2, 1);
      cnt++;
    }
    long long pos = b[idx];
    l = 1;
    r = n;
    ans = 0;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (query(mid) < pos) {
        l = mid + 1;
        ans = mid;
      } else {
        r = mid - 1;
      }
    }
    ans++;
    dp[idx] = ans;
  }
  for (i = 1; i <= (long long)(m); i++) cout << a[dp[i]] << '\n';
  return 0;
}

```