#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long mo = 1e9 + 7;
long long sx, sy, ex, ey, dx[6] = {0, 1, 0, -1, 0, 0},
                          dy[6] = {1, 0, -1, 0, 0, 0}, m, n, k,
                          dz[6]{0, 0, 0, 0, -1, 1}, sg;
long long p, no, v, ans, w;
int par[550000];
long long a[500005], b[400005], c[500006], d1[400006], ma[500006], mi[500006];
int lg[200005], mi2[50], mn[25][200005];
long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
set<int> se;
struct node {
  long long u, v, w;
} a2[200005];
vector<node> eg;
long long qu(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans % m * a % m;
    }
    b >>= 1;
    a = a % m * a % m;
  }
  return ans;
}
int su(int n) {
  if (n == 1 || n == 0) return 0;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int fi(int a) {
  if (a == par[a])
    return a;
  else
    return par[a] = fi(par[a]);
}
map<int, long long> mp, mp1;
map<pair<long long, int>, int> mp2;
priority_queue<pair<long long, int> > que;
pair<long long, int> a1[400000];
int vis[400055], vis1[400055], bit[200005];
long long dp[55][1005];
char maze[505][505];
vector<int> g[400000], g1[400000];
vector<pair<long long, int> > ve, ve1, ve2;
int cmp1(pair<long long, int> a, pair<long long, int> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
int cmp(node a, node b) {
  if (a.u == b.u) return a.v < b.v;
  return a.u < b.u;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
string ans1;
int query(int i) {
  long long ans = 0;
  while (i > 0) {
    ans += bit[i];
    i -= i & -i;
  }
  return ans;
}
void add(int i, int x) {
  while (i <= n) {
    bit[i] += x;
    i += i & -i;
  }
}
int main() {
  int t, p2, p3;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, ss, sss;
  long long l, r, n1, u, l1, r1;
  int tot = 1;
  char ch;
  while (cin >> n) {
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      a1[i] = pair<long long, int>{a[i], i};
    }
    sort(a1 + 1, a1 + n + 1, cmp1);
    cin >> m;
    for (int i = 1; i <= m; i++) {
      cin >> p >> w;
      a2[i] = node{p, w, i};
    }
    sort(a2 + 1, a2 + m + 1, cmp);
    no = 0;
    for (int i = 1; i <= m; i++) {
      while (no < a2[i].u) {
        no++;
        add(a1[no].second, 1);
      }
      l = 0, r = n + 1;
      while (r - l > 1) {
        int mid = (l + r) / 2;
        if (query(mid) >= a2[i].v)
          r = mid;
        else
          l = mid;
      }
      vis[a2[i].w] = a[r];
    }
    for (int i = 1; i <= m; i++) cout << vis[i] << endl;
  }
}

### Prompt

Please provide a CPP coded solution 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>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long mo = 1e9 + 7;
long long sx, sy, ex, ey, dx[6] = {0, 1, 0, -1, 0, 0},
                          dy[6] = {1, 0, -1, 0, 0, 0}, m, n, k,
                          dz[6]{0, 0, 0, 0, -1, 1}, sg;
long long p, no, v, ans, w;
int par[550000];
long long a[500005], b[400005], c[500006], d1[400006], ma[500006], mi[500006];
int lg[200005], mi2[50], mn[25][200005];
long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
set<int> se;
struct node {
  long long u, v, w;
} a2[200005];
vector<node> eg;
long long qu(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans % m * a % m;
    }
    b >>= 1;
    a = a % m * a % m;
  }
  return ans;
}
int su(int n) {
  if (n == 1 || n == 0) return 0;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int fi(int a) {
  if (a == par[a])
    return a;
  else
    return par[a] = fi(par[a]);
}
map<int, long long> mp, mp1;
map<pair<long long, int>, int> mp2;
priority_queue<pair<long long, int> > que;
pair<long long, int> a1[400000];
int vis[400055], vis1[400055], bit[200005];
long long dp[55][1005];
char maze[505][505];
vector<int> g[400000], g1[400000];
vector<pair<long long, int> > ve, ve1, ve2;
int cmp1(pair<long long, int> a, pair<long long, int> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
int cmp(node a, node b) {
  if (a.u == b.u) return a.v < b.v;
  return a.u < b.u;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
string ans1;
int query(int i) {
  long long ans = 0;
  while (i > 0) {
    ans += bit[i];
    i -= i & -i;
  }
  return ans;
}
void add(int i, int x) {
  while (i <= n) {
    bit[i] += x;
    i += i & -i;
  }
}
int main() {
  int t, p2, p3;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, ss, sss;
  long long l, r, n1, u, l1, r1;
  int tot = 1;
  char ch;
  while (cin >> n) {
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      a1[i] = pair<long long, int>{a[i], i};
    }
    sort(a1 + 1, a1 + n + 1, cmp1);
    cin >> m;
    for (int i = 1; i <= m; i++) {
      cin >> p >> w;
      a2[i] = node{p, w, i};
    }
    sort(a2 + 1, a2 + m + 1, cmp);
    no = 0;
    for (int i = 1; i <= m; i++) {
      while (no < a2[i].u) {
        no++;
        add(a1[no].second, 1);
      }
      l = 0, r = n + 1;
      while (r - l > 1) {
        int mid = (l + r) / 2;
        if (query(mid) >= a2[i].v)
          r = mid;
        else
          l = mid;
      }
      vis[a2[i].w] = a[r];
    }
    for (int i = 1; i <= m; i++) cout << vis[i] << endl;
  }
}

```