#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 2e5 + 100;
int n, a[N], m, ans[N];
vector<pair<int, int>> v;
struct query {
  int k, pos, id;
  bool operator<(query o) const { return k < o.k; }
} Q[N];
const int RTE = 2e9;
struct ds {
  struct node {
    int val, prior, size;
    node *l, *r;
  };
  typedef node *pnode;
  pnode root;
  ds() { root = NULL; }
  int sz(pnode t) { return t ? t->size : 0; }
  void upd_sz(pnode t) {
    if (t) t->size = sz(t->l) + 1 + sz(t->r);
  }
  pnode init(int x) {
    pnode ret = (pnode)malloc(sizeof(node));
    ret->val = x, ret->prior = rand(), ret->size = 1;
    ret->l = ret->r = NULL;
    return ret;
  }
  void split(pnode t, pnode &l, pnode &r, int key) {
    if (!t)
      l = r = NULL;
    else if (t->val <= key)
      split(t->r, t->r, r, key), l = t;
    else
      split(t->l, l, t->l, key), r = t;
    upd_sz(t);
  }
  void merge(pnode &t, pnode l, pnode r) {
    if (!l || !r)
      t = l ? l : r;
    else if (l->prior > r->prior)
      merge(l->r, l->r, r), t = l;
    else
      merge(r->l, l, r->l), t = r;
    upd_sz(t);
  }
  bool find(pnode t, int key) {
    if (!t) return false;
    if (t->val == key) return true;
    return find(t->val < key ? t->r : t->l, key);
  }
  void insert(pnode it, pnode &t) {
    if (!t)
      t = it;
    else if (it->prior > t->prior)
      split(t, it->l, it->r, it->val), t = it;
    else
      insert(it, t->val <= it->val ? t->r : t->l);
    upd_sz(t);
  }
  void insert(int val) { insert(init(val), root); }
  void erase(int key, pnode &t) {
    if (!t) return;
    if (t->val == key) {
      pnode temp = t;
      merge(t, t->l, t->r);
      free(temp);
    } else
      erase(key, t->val < key ? t->r : t->l);
    upd_sz(t);
  }
  void erase(int key) { erase(key, root); }
  int kth(int k, pnode t) {
    if (sz(t->l) == k - 1) return t->val;
    if (sz(t->l) >= k) return kth(k, t->l);
    return kth(k - 1 - sz(t->l), t->r);
  }
  int kth(int k) { return kth(k, root); }
  int count(int s, pnode t) {
    if (!t) return 0;
    if (t->val >= s) return count(s, t->l);
    return sz(t->l) + 1 + count(s, t->r);
  }
  int count(int s) { return count(s, root); }
} treap;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    v.push_back({-a[i], i});
  }
  sort(v.begin(), v.end());
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &Q[i].k, &Q[i].pos);
    Q[i].id = i;
  }
  sort(Q, Q + m);
  int cur = 0;
  for (int i = 0; i < m; i++) {
    while (cur < Q[i].k) {
      treap.insert(v[cur].second);
      cur++;
    }
    ans[Q[i].id] = a[treap.kth(Q[i].pos)];
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
using ll = long long;
const int N = 2e5 + 100;
int n, a[N], m, ans[N];
vector<pair<int, int>> v;
struct query {
  int k, pos, id;
  bool operator<(query o) const { return k < o.k; }
} Q[N];
const int RTE = 2e9;
struct ds {
  struct node {
    int val, prior, size;
    node *l, *r;
  };
  typedef node *pnode;
  pnode root;
  ds() { root = NULL; }
  int sz(pnode t) { return t ? t->size : 0; }
  void upd_sz(pnode t) {
    if (t) t->size = sz(t->l) + 1 + sz(t->r);
  }
  pnode init(int x) {
    pnode ret = (pnode)malloc(sizeof(node));
    ret->val = x, ret->prior = rand(), ret->size = 1;
    ret->l = ret->r = NULL;
    return ret;
  }
  void split(pnode t, pnode &l, pnode &r, int key) {
    if (!t)
      l = r = NULL;
    else if (t->val <= key)
      split(t->r, t->r, r, key), l = t;
    else
      split(t->l, l, t->l, key), r = t;
    upd_sz(t);
  }
  void merge(pnode &t, pnode l, pnode r) {
    if (!l || !r)
      t = l ? l : r;
    else if (l->prior > r->prior)
      merge(l->r, l->r, r), t = l;
    else
      merge(r->l, l, r->l), t = r;
    upd_sz(t);
  }
  bool find(pnode t, int key) {
    if (!t) return false;
    if (t->val == key) return true;
    return find(t->val < key ? t->r : t->l, key);
  }
  void insert(pnode it, pnode &t) {
    if (!t)
      t = it;
    else if (it->prior > t->prior)
      split(t, it->l, it->r, it->val), t = it;
    else
      insert(it, t->val <= it->val ? t->r : t->l);
    upd_sz(t);
  }
  void insert(int val) { insert(init(val), root); }
  void erase(int key, pnode &t) {
    if (!t) return;
    if (t->val == key) {
      pnode temp = t;
      merge(t, t->l, t->r);
      free(temp);
    } else
      erase(key, t->val < key ? t->r : t->l);
    upd_sz(t);
  }
  void erase(int key) { erase(key, root); }
  int kth(int k, pnode t) {
    if (sz(t->l) == k - 1) return t->val;
    if (sz(t->l) >= k) return kth(k, t->l);
    return kth(k - 1 - sz(t->l), t->r);
  }
  int kth(int k) { return kth(k, root); }
  int count(int s, pnode t) {
    if (!t) return 0;
    if (t->val >= s) return count(s, t->l);
    return sz(t->l) + 1 + count(s, t->r);
  }
  int count(int s) { return count(s, root); }
} treap;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    v.push_back({-a[i], i});
  }
  sort(v.begin(), v.end());
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &Q[i].k, &Q[i].pos);
    Q[i].id = i;
  }
  sort(Q, Q + m);
  int cur = 0;
  for (int i = 0; i < m; i++) {
    while (cur < Q[i].k) {
      treap.insert(v[cur].second);
      cur++;
    }
    ans[Q[i].id] = a[treap.kth(Q[i].pos)];
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}

```