#include <bits/stdc++.h>
using namespace std;
const int INF = (1 << 29);
const int INFF = 0x7fffffff;
const long long LINF = (1ll << 60);
const long long LINFF = 0x7fffffffffffffff;
const double PI = 3.14159265359;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rnd(long long a, long long b) {
  return uniform_int_distribution<long long>(a, b)(rng);
}
template <typename T>
struct treap {
 private:
  struct node {
    T key;
    int prior;
    node *l = nullptr, *r = nullptr;
    int cnt = 1;
    node(T k) : key(k), prior(rand()) {}
  };
  node *t = nullptr;
  int cnt(node *n) { return n ? n->cnt : 0; }
  void upd_cnt(node *n) {
    if (n) n->cnt = 1 + cnt(n->l) + cnt(n->r);
  }
  void split(node *root, T key, node *&l, node *&r) {
    if (!root)
      l = r = nullptr;
    else if (key <= root->key) {
      r = root;
      split(root->l, key, l, root->l);
    } else {
      l = root;
      split(root->r, key, root->r, r);
    }
    upd_cnt(l);
    upd_cnt(r);
  }
  void insert(node *&root, node *element) {
    if (!root) return void(root = element);
    if (element->prior > root->prior) {
      split(root, element->key, element->l, element->r);
      root = element;
    } else {
      if (element->key < root->key)
        insert(root->l, element);
      else
        insert(root->r, element);
    }
    upd_cnt(root);
  }
  void merge(node *&root, node *l, node *r) {
    if (!r || !l)
      root = l ? l : r;
    else if (l->prior > r->prior) {
      root = l;
      merge(l->r, l->r, r);
    } else {
      root = r;
      merge(r->l, l, r->l);
    }
    upd_cnt(root);
  }
  void erase(node *&root, T key) {
    if (!root) return;
    if (root->key == key) {
      node *aux = root;
      merge(root, root->l, root->r);
      delete aux;
    } else {
      if (key < root->key)
        erase(root->l, key);
      else
        erase(root->r, key);
    }
    upd_cnt(root);
  }
  void del(node *n) {
    if (!n) return;
    del(n->l);
    del(n->r);
    delete n;
  }
  T kth(node *root, int pos) {
    if (!root) return {0, 0};
    int p = 1 + cnt(root->l);
    if (p == pos) return root->key;
    if (p > pos)
      return kth(root->l, pos);
    else
      return kth(root->r, pos - p);
  }

 public:
  void insert(T key) { insert(t, new node(key)); }
  void erase(T key) { erase(t, key); }
  ~treap() { del(t); }
  T kth(int pos) { return kth(t, pos + 1); }
  int size() { return cnt(t); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  vector<pair<int, int>> vv;
  for (int i = 0; i < n; i++) vv.push_back({i, v[i]});
  sort(vv.begin(), vv.end(),
       [](const pair<int, int> &a, const pair<int, int> &b) {
         if (a.second == b.second)
           return a.first < b.first;
         else
           return a.second > b.second;
       });
  vector<tuple<int, int, int>> queries;
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int k, pos;
    cin >> k >> pos;
    queries.push_back({k, pos, i});
  }
  sort(queries.begin(), queries.end());
  vector<int> ans(q);
  treap<pair<int, int>> trp;
  int j = 0;
  for (int i = 0; i < q; i++) {
    int k, pos, id;
    tie(k, pos, id) = queries[i];
    while (trp.size() < k) trp.insert(vv[j++]);
    ans[id] = trp.kth(pos - 1).second;
  }
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int INF = (1 << 29);
const int INFF = 0x7fffffff;
const long long LINF = (1ll << 60);
const long long LINFF = 0x7fffffffffffffff;
const double PI = 3.14159265359;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rnd(long long a, long long b) {
  return uniform_int_distribution<long long>(a, b)(rng);
}
template <typename T>
struct treap {
 private:
  struct node {
    T key;
    int prior;
    node *l = nullptr, *r = nullptr;
    int cnt = 1;
    node(T k) : key(k), prior(rand()) {}
  };
  node *t = nullptr;
  int cnt(node *n) { return n ? n->cnt : 0; }
  void upd_cnt(node *n) {
    if (n) n->cnt = 1 + cnt(n->l) + cnt(n->r);
  }
  void split(node *root, T key, node *&l, node *&r) {
    if (!root)
      l = r = nullptr;
    else if (key <= root->key) {
      r = root;
      split(root->l, key, l, root->l);
    } else {
      l = root;
      split(root->r, key, root->r, r);
    }
    upd_cnt(l);
    upd_cnt(r);
  }
  void insert(node *&root, node *element) {
    if (!root) return void(root = element);
    if (element->prior > root->prior) {
      split(root, element->key, element->l, element->r);
      root = element;
    } else {
      if (element->key < root->key)
        insert(root->l, element);
      else
        insert(root->r, element);
    }
    upd_cnt(root);
  }
  void merge(node *&root, node *l, node *r) {
    if (!r || !l)
      root = l ? l : r;
    else if (l->prior > r->prior) {
      root = l;
      merge(l->r, l->r, r);
    } else {
      root = r;
      merge(r->l, l, r->l);
    }
    upd_cnt(root);
  }
  void erase(node *&root, T key) {
    if (!root) return;
    if (root->key == key) {
      node *aux = root;
      merge(root, root->l, root->r);
      delete aux;
    } else {
      if (key < root->key)
        erase(root->l, key);
      else
        erase(root->r, key);
    }
    upd_cnt(root);
  }
  void del(node *n) {
    if (!n) return;
    del(n->l);
    del(n->r);
    delete n;
  }
  T kth(node *root, int pos) {
    if (!root) return {0, 0};
    int p = 1 + cnt(root->l);
    if (p == pos) return root->key;
    if (p > pos)
      return kth(root->l, pos);
    else
      return kth(root->r, pos - p);
  }

 public:
  void insert(T key) { insert(t, new node(key)); }
  void erase(T key) { erase(t, key); }
  ~treap() { del(t); }
  T kth(int pos) { return kth(t, pos + 1); }
  int size() { return cnt(t); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  vector<pair<int, int>> vv;
  for (int i = 0; i < n; i++) vv.push_back({i, v[i]});
  sort(vv.begin(), vv.end(),
       [](const pair<int, int> &a, const pair<int, int> &b) {
         if (a.second == b.second)
           return a.first < b.first;
         else
           return a.second > b.second;
       });
  vector<tuple<int, int, int>> queries;
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int k, pos;
    cin >> k >> pos;
    queries.push_back({k, pos, i});
  }
  sort(queries.begin(), queries.end());
  vector<int> ans(q);
  treap<pair<int, int>> trp;
  int j = 0;
  for (int i = 0; i < q; i++) {
    int k, pos, id;
    tie(k, pos, id) = queries[i];
    while (trp.size() < k) trp.insert(vv[j++]);
    ans[id] = trp.kth(pos - 1).second;
  }
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
  return 0;
}

```