#include <bits/stdc++.h>
using namespace std;
int n, m, k, pos;
vector<tuple<int, int, int>> t;
map<int, set<int>> mp;
unordered_map<int, int> u;
struct node {
  int x, y, s;
  node *left, *right;
  node(int val) {
    x = val;
    y = rand();
    s = 1;
    left = nullptr;
    right = nullptr;
  }
};
int get_size(node *root) {
  if (!root) return 0;
  return root->s;
}
void update(node *root) {
  root->s = 1 + get_size(root->left) + get_size(root->right);
}
pair<node *, node *> split(node *root, int val) {
  if (!root) return {nullptr, nullptr};
  if (root->x <= val) {
    pair<node *, node *> res = split(root->right, val);
    root->right = res.first;
    update(root);
    return {root, res.second};
  } else {
    pair<node *, node *> res = split(root->left, val);
    root->left = res.second;
    update(root);
    return {res.first, root};
  }
}
node *merg(node *root1, node *root2) {
  if (!root1) return root2;
  if (!root2) return root1;
  if (root1->y < root2->y) {
    root1->right = merg(root1->right, root2);
    update(root1);
    return root1;
  } else {
    root2->left = merg(root1, root2->left);
    update(root2);
    return root2;
  }
}
int cut(node *root, int val) {
  if (get_size(root->left) >= val)
    return cut(root->left, val);
  else if (get_size(root->left) + 1 == val)
    return root->x;
  else
    return cut(root->right, val - get_size(root->left) - 1);
}
int main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  vector<int> a(n), ara, g;
  node *root = nullptr;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    g.push_back(a[i]);
    mp[a[i]].insert(i);
  }
  sort(g.begin(), g.end(), greater<int>());
  for (int x : g) {
    if (u[x] > 0) continue;
    u[x] = 1;
    for (int y : mp[x]) ara.push_back(y);
  }
  cin >> m;
  vector<int> ans(m);
  for (int j = 0; j < m; j++) {
    cin >> k >> pos;
    t.push_back(make_tuple(k, pos, j));
  }
  sort(t.begin(), t.end());
  int i = 0;
  for (int j = 0; j < m; j++) {
    int num;
    tie(k, pos, num) = t[j];
    while (i < k) {
      pair<node *, node *> res = split(root, ara[i]);
      root = merg(merg(res.first, new node(ara[i])), res.second);
      i++;
    }
    ans[num] = a[cut(root, pos)];
  }
  for (int j = 0; j < m; j++) cout << ans[j] << "\n";
  return 0;
}

### Prompt

Please formulate a Cpp solution to 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 n, m, k, pos;
vector<tuple<int, int, int>> t;
map<int, set<int>> mp;
unordered_map<int, int> u;
struct node {
  int x, y, s;
  node *left, *right;
  node(int val) {
    x = val;
    y = rand();
    s = 1;
    left = nullptr;
    right = nullptr;
  }
};
int get_size(node *root) {
  if (!root) return 0;
  return root->s;
}
void update(node *root) {
  root->s = 1 + get_size(root->left) + get_size(root->right);
}
pair<node *, node *> split(node *root, int val) {
  if (!root) return {nullptr, nullptr};
  if (root->x <= val) {
    pair<node *, node *> res = split(root->right, val);
    root->right = res.first;
    update(root);
    return {root, res.second};
  } else {
    pair<node *, node *> res = split(root->left, val);
    root->left = res.second;
    update(root);
    return {res.first, root};
  }
}
node *merg(node *root1, node *root2) {
  if (!root1) return root2;
  if (!root2) return root1;
  if (root1->y < root2->y) {
    root1->right = merg(root1->right, root2);
    update(root1);
    return root1;
  } else {
    root2->left = merg(root1, root2->left);
    update(root2);
    return root2;
  }
}
int cut(node *root, int val) {
  if (get_size(root->left) >= val)
    return cut(root->left, val);
  else if (get_size(root->left) + 1 == val)
    return root->x;
  else
    return cut(root->right, val - get_size(root->left) - 1);
}
int main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  vector<int> a(n), ara, g;
  node *root = nullptr;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    g.push_back(a[i]);
    mp[a[i]].insert(i);
  }
  sort(g.begin(), g.end(), greater<int>());
  for (int x : g) {
    if (u[x] > 0) continue;
    u[x] = 1;
    for (int y : mp[x]) ara.push_back(y);
  }
  cin >> m;
  vector<int> ans(m);
  for (int j = 0; j < m; j++) {
    cin >> k >> pos;
    t.push_back(make_tuple(k, pos, j));
  }
  sort(t.begin(), t.end());
  int i = 0;
  for (int j = 0; j < m; j++) {
    int num;
    tie(k, pos, num) = t[j];
    while (i < k) {
      pair<node *, node *> res = split(root, ara[i]);
      root = merg(merg(res.first, new node(ara[i])), res.second);
      i++;
    }
    ans[num] = a[cut(root, pos)];
  }
  for (int j = 0; j < m; j++) cout << ans[j] << "\n";
  return 0;
}

```