#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(42);
const int inf = 1000000000;
const int MAXSZ = 2e5 + 10;
int n, m, ar[MAXSZ], ans[MAXSZ];
vector<pair<int, int> > reqs[MAXSZ];
struct node {
  int val, cnt, pr;
  node *l = nullptr, *r = nullptr;
  node() {}
  node(int x) : val(x), cnt(1), pr(rnd()) {}
};
node *treap;
inline int getcnt(node *r) { return (r ? r->cnt : 0); }
inline void update(node *&x) {
  if (x) x->cnt = 1 + getcnt(x->l) + getcnt(x->r);
}
node *merge(node *tl, node *tr) {
  if (!tl || !tr) return tl ? tl : tr;
  if (tl->pr > tr->pr) {
    tl->r = merge(tl->r, tr), update(tl);
    return tl;
  } else {
    tr->l = merge(tl, tr->l), update(tr);
    return tr;
  }
}
inline void addnew(int val) {
  node *newnode = new node(val);
  treap = merge(treap, newnode);
}
pair<node *, node *> split(node *t, int val) {
  if (!t) return {nullptr, nullptr};
  if (getcnt(t->l) >= val) {
    pair<node *, node *> res = split(t->l, val);
    t->l = res.second, update(t);
    return {res.first, t};
  } else {
    pair<node *, node *> res = split(t->r, val - getcnt(t->l) - 1);
    t->r = res.first, update(t);
    return {t, res.second};
  }
}
void remove(int val, node *&t = treap) {
  if (!t) return;
  if (t->val == val) {
    t = merge(t->l, t->r);
    return;
  }
  if (val < t->val)
    remove(val, t->l);
  else
    remove(val, t->r);
  update(t);
}
int getans(int val) {
  pair<node *, node *> tmp = split(treap, val);
  pair<node *, node *> t = split(tmp.second, 1);
  int ret = t.first->val;
  treap = merge(tmp.first, merge(t.first, t.second));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ar[i];
  cin >> m;
  for (int i = 0; i < m; i++) {
    int t1, t2;
    cin >> t1 >> t2;
    reqs[t1].push_back({t2, i});
  }
  set<pair<int, int> > cur;
  for (int i = 0; i < n; i++) cur.insert({ar[i], -i}), addnew(i);
  for (int i = n; i > 0; i--) {
    for (auto tmp : reqs[i]) {
      ans[tmp.second] = getans(tmp.first - 1);
    }
    int tmp = -cur.begin()->second;
    cur.erase(cur.begin());
    remove(tmp);
  }
  for (int i = 0; i < m; i++) cout << ar[ans[i]] << endl;
  return 0;
}

### Prompt

Your task is to create 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;
mt19937 rnd(42);
const int inf = 1000000000;
const int MAXSZ = 2e5 + 10;
int n, m, ar[MAXSZ], ans[MAXSZ];
vector<pair<int, int> > reqs[MAXSZ];
struct node {
  int val, cnt, pr;
  node *l = nullptr, *r = nullptr;
  node() {}
  node(int x) : val(x), cnt(1), pr(rnd()) {}
};
node *treap;
inline int getcnt(node *r) { return (r ? r->cnt : 0); }
inline void update(node *&x) {
  if (x) x->cnt = 1 + getcnt(x->l) + getcnt(x->r);
}
node *merge(node *tl, node *tr) {
  if (!tl || !tr) return tl ? tl : tr;
  if (tl->pr > tr->pr) {
    tl->r = merge(tl->r, tr), update(tl);
    return tl;
  } else {
    tr->l = merge(tl, tr->l), update(tr);
    return tr;
  }
}
inline void addnew(int val) {
  node *newnode = new node(val);
  treap = merge(treap, newnode);
}
pair<node *, node *> split(node *t, int val) {
  if (!t) return {nullptr, nullptr};
  if (getcnt(t->l) >= val) {
    pair<node *, node *> res = split(t->l, val);
    t->l = res.second, update(t);
    return {res.first, t};
  } else {
    pair<node *, node *> res = split(t->r, val - getcnt(t->l) - 1);
    t->r = res.first, update(t);
    return {t, res.second};
  }
}
void remove(int val, node *&t = treap) {
  if (!t) return;
  if (t->val == val) {
    t = merge(t->l, t->r);
    return;
  }
  if (val < t->val)
    remove(val, t->l);
  else
    remove(val, t->r);
  update(t);
}
int getans(int val) {
  pair<node *, node *> tmp = split(treap, val);
  pair<node *, node *> t = split(tmp.second, 1);
  int ret = t.first->val;
  treap = merge(tmp.first, merge(t.first, t.second));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ar[i];
  cin >> m;
  for (int i = 0; i < m; i++) {
    int t1, t2;
    cin >> t1 >> t2;
    reqs[t1].push_back({t2, i});
  }
  set<pair<int, int> > cur;
  for (int i = 0; i < n; i++) cur.insert({ar[i], -i}), addnew(i);
  for (int i = n; i > 0; i--) {
    for (auto tmp : reqs[i]) {
      ans[tmp.second] = getans(tmp.first - 1);
    }
    int tmp = -cur.begin()->second;
    cur.erase(cur.begin());
    remove(tmp);
  }
  for (int i = 0; i < m; i++) cout << ar[ans[i]] << endl;
  return 0;
}

```