#include <bits/stdc++.h>
using namespace std;
int niz[400005];
int seg[1000005];
int poz[1000005];
void init(int node, int l, int r) {
  if (l == r) {
    seg[node] = niz[l];
    poz[node] = l;
    return;
  }
  int mid = (l + r) / 2;
  init(node * 2, l, mid);
  init(node * 2 + 1, mid + 1, r);
  seg[node] = max(seg[node * 2], seg[node * 2 + 1]);
  if (seg[node * 2] >= seg[node * 2 + 1])
    poz[node] = poz[node * 2];
  else
    poz[node] = poz[node * 2 + 1];
}
void update(int node, int l, int r, int pos) {
  if (l == r) {
    seg[node] = -1e9;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(node * 2, l, mid, pos);
  else
    update(node * 2 + 1, mid + 1, r, pos);
  seg[node] = max(seg[node * 2], seg[node * 2 + 1]);
  if (seg[node * 2] >= seg[node * 2 + 1])
    poz[node] = poz[node * 2];
  else
    poz[node] = poz[node * 2 + 1];
}
int n;
int m;
struct kv {
  int x, y, z;
};
vector<kv> upit;
bool cmp(kv a, kv b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
int ans[400005];
int bit[400005];
void upd(int x) {
  while (x <= 200000) {
    bit[x]++;
    x += (x & -x);
  }
}
int query(int x) {
  int res = 0;
  while (x > 0) {
    res += bit[x];
    x -= (x & -x);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cerr.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> niz[i];
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    upit.push_back({a, b, i});
  }
  sort(upit.begin(), upit.end(), cmp);
  int br = 0;
  init(1, 1, n);
  for (int i = 1; i <= n; i++) {
    if (br >= m) break;
    int koji = poz[1];
    update(1, 1, n, koji);
    upd(koji);
    while (upit[br].x < i && br < m) ++br;
    if (br >= m) break;
    while (upit[br].x == i) {
      int l = 1;
      int r = n;
      while (l < r) {
        int mid = (l + r) / 2;
        if (query(mid) >= upit[br].y)
          r = mid;
        else
          l = mid + 1;
      }
      ans[upit[br].z] = l;
      ++br;
    }
  }
  for (int i = 1; i <= m; i++) cout << niz[ans[i]] << endl;
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock 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 ≤ 100) — 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 ≤ 100) — 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;
int niz[400005];
int seg[1000005];
int poz[1000005];
void init(int node, int l, int r) {
  if (l == r) {
    seg[node] = niz[l];
    poz[node] = l;
    return;
  }
  int mid = (l + r) / 2;
  init(node * 2, l, mid);
  init(node * 2 + 1, mid + 1, r);
  seg[node] = max(seg[node * 2], seg[node * 2 + 1]);
  if (seg[node * 2] >= seg[node * 2 + 1])
    poz[node] = poz[node * 2];
  else
    poz[node] = poz[node * 2 + 1];
}
void update(int node, int l, int r, int pos) {
  if (l == r) {
    seg[node] = -1e9;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(node * 2, l, mid, pos);
  else
    update(node * 2 + 1, mid + 1, r, pos);
  seg[node] = max(seg[node * 2], seg[node * 2 + 1]);
  if (seg[node * 2] >= seg[node * 2 + 1])
    poz[node] = poz[node * 2];
  else
    poz[node] = poz[node * 2 + 1];
}
int n;
int m;
struct kv {
  int x, y, z;
};
vector<kv> upit;
bool cmp(kv a, kv b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
int ans[400005];
int bit[400005];
void upd(int x) {
  while (x <= 200000) {
    bit[x]++;
    x += (x & -x);
  }
}
int query(int x) {
  int res = 0;
  while (x > 0) {
    res += bit[x];
    x -= (x & -x);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cerr.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> niz[i];
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    upit.push_back({a, b, i});
  }
  sort(upit.begin(), upit.end(), cmp);
  int br = 0;
  init(1, 1, n);
  for (int i = 1; i <= n; i++) {
    if (br >= m) break;
    int koji = poz[1];
    update(1, 1, n, koji);
    upd(koji);
    while (upit[br].x < i && br < m) ++br;
    if (br >= m) break;
    while (upit[br].x == i) {
      int l = 1;
      int r = n;
      while (l < r) {
        int mid = (l + r) / 2;
        if (query(mid) >= upit[br].y)
          r = mid;
        else
          l = mid + 1;
      }
      ans[upit[br].z] = l;
      ++br;
    }
  }
  for (int i = 1; i <= m; i++) cout << niz[ans[i]] << endl;
  return 0;
}

```