#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 20;
const long long INF = INT_MAX;
struct node {
  node *l, *r;
  long long x, y;
  long long sum;
  long long val;
  long long sz;
  node(long long s, long long pos) {
    l = nullptr;
    r = nullptr;
    y = rand();
    sum = s * s;
    val = s;
    x = pos;
    sz = 1;
  }
};
long long getx(node* v) {
  if (v == nullptr) return 0;
  return v->x;
}
long long getsum(node* v) {
  if (v == nullptr) return 0;
  return v->sum;
}
long long getval(node* v) {
  if (v == nullptr) return 0;
  return v->val;
}
long long getsz(node* v) {
  if (v == nullptr) return 0;
  return v->sz;
}
void add_val(node*& v, long long newv) {
  if (v == nullptr) return;
  v->val += newv;
  v->sum = getval(v) * getval(v);
  v->sum += getsum(v->l) + getsum(v->r);
}
void change_val(node*& v, long long newv) {
  if (v == nullptr) return;
  v->val = newv;
  v->sum = newv * newv;
  v->sum += getsum(v->l) + getsum(v->r);
}
void upd(node* v) {
  if (v == nullptr) return;
  v->sum = getsum(v->l) + getsum(v->r) + getval(v) * getval(v);
  v->sz = getsz(v->l) + getsz(v->r) + 1;
}
pair<node*, node*> split(node* root, long long key) {
  if (root == nullptr) return {nullptr, nullptr};
  if (root->x <= key) {
    pair<node*, node*> splitted = split(root->r, key);
    root->r = splitted.first;
    upd(root);
    return {root, splitted.second};
  } else {
    pair<node*, node*> splitted = split(root->l, key);
    root->l = splitted.second;
    upd(root);
    return {splitted.first, root};
  }
}
node* merge(node* left, node* right) {
  if (left == nullptr || right == nullptr) {
    return right != nullptr ? right : left;
  }
  if (left->y > right->y) {
    left->r = merge(left->r, right);
    upd(left);
    return left;
  } else {
    right->l = merge(left, right->l);
    upd(right);
    return right;
  }
}
void insert(node*& root, long long key, long long val) {
  auto splitted = split(root, key - 1);
  root = merge(merge(splitted.first, new node(val, key)), splitted.second);
}
long long find(node*& root, long long n) {
  long long ln = getsz(root->l);
  if (ln + 1 == n) {
    return getval(root);
  }
  if (n < ln + 1) {
    return find(root->l, n);
  } else
    return find(root->r, n - ln - 1);
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a;
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    a.push_back({x, i});
  }
  sort(a.begin(), a.end(), cmp);
  long long t;
  cin >> t;
  vector<pair<pair<long long, long long>, long long>> q;
  for (long long i = 0; i < t; ++i) {
    long long k, pos;
    cin >> k >> pos;
    q.push_back({{k, pos}, i});
  }
  sort(q.begin(), q.end());
  node* root = nullptr;
  long long k = 0;
  vector<long long> qqq(q.size());
  for (auto& i : q) {
    long long curk = i.first.first;
    long long pos = i.first.second;
    long long id = i.second;
    while (k < curk) {
      insert(root, a[k].second + 1, a[k].first);
      k++;
    }
    qqq[id] = find(root, pos);
  }
  for (auto& i : qqq) cout << i << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const long long N = 2e5 + 20;
const long long INF = INT_MAX;
struct node {
  node *l, *r;
  long long x, y;
  long long sum;
  long long val;
  long long sz;
  node(long long s, long long pos) {
    l = nullptr;
    r = nullptr;
    y = rand();
    sum = s * s;
    val = s;
    x = pos;
    sz = 1;
  }
};
long long getx(node* v) {
  if (v == nullptr) return 0;
  return v->x;
}
long long getsum(node* v) {
  if (v == nullptr) return 0;
  return v->sum;
}
long long getval(node* v) {
  if (v == nullptr) return 0;
  return v->val;
}
long long getsz(node* v) {
  if (v == nullptr) return 0;
  return v->sz;
}
void add_val(node*& v, long long newv) {
  if (v == nullptr) return;
  v->val += newv;
  v->sum = getval(v) * getval(v);
  v->sum += getsum(v->l) + getsum(v->r);
}
void change_val(node*& v, long long newv) {
  if (v == nullptr) return;
  v->val = newv;
  v->sum = newv * newv;
  v->sum += getsum(v->l) + getsum(v->r);
}
void upd(node* v) {
  if (v == nullptr) return;
  v->sum = getsum(v->l) + getsum(v->r) + getval(v) * getval(v);
  v->sz = getsz(v->l) + getsz(v->r) + 1;
}
pair<node*, node*> split(node* root, long long key) {
  if (root == nullptr) return {nullptr, nullptr};
  if (root->x <= key) {
    pair<node*, node*> splitted = split(root->r, key);
    root->r = splitted.first;
    upd(root);
    return {root, splitted.second};
  } else {
    pair<node*, node*> splitted = split(root->l, key);
    root->l = splitted.second;
    upd(root);
    return {splitted.first, root};
  }
}
node* merge(node* left, node* right) {
  if (left == nullptr || right == nullptr) {
    return right != nullptr ? right : left;
  }
  if (left->y > right->y) {
    left->r = merge(left->r, right);
    upd(left);
    return left;
  } else {
    right->l = merge(left, right->l);
    upd(right);
    return right;
  }
}
void insert(node*& root, long long key, long long val) {
  auto splitted = split(root, key - 1);
  root = merge(merge(splitted.first, new node(val, key)), splitted.second);
}
long long find(node*& root, long long n) {
  long long ln = getsz(root->l);
  if (ln + 1 == n) {
    return getval(root);
  }
  if (n < ln + 1) {
    return find(root->l, n);
  } else
    return find(root->r, n - ln - 1);
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a;
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    a.push_back({x, i});
  }
  sort(a.begin(), a.end(), cmp);
  long long t;
  cin >> t;
  vector<pair<pair<long long, long long>, long long>> q;
  for (long long i = 0; i < t; ++i) {
    long long k, pos;
    cin >> k >> pos;
    q.push_back({{k, pos}, i});
  }
  sort(q.begin(), q.end());
  node* root = nullptr;
  long long k = 0;
  vector<long long> qqq(q.size());
  for (auto& i : q) {
    long long curk = i.first.first;
    long long pos = i.first.second;
    long long id = i.second;
    while (k < curk) {
      insert(root, a[k].second + 1, a[k].first);
      k++;
    }
    qqq[id] = find(root, pos);
  }
  for (auto& i : qqq) cout << i << endl;
  return 0;
}

```