#include <bits/stdc++.h>
using namespace std;
int a[200020];
priority_queue<pair<int, int> > pq;
vector<pair<int, pair<int, int> > > v;
int k, pos;
int ans[200020];
vector<int> arr;
class Node {
 public:
  int key;
  Node *left;
  Node *right;
  int height;
  int num;
};
int get_num(Node *N) {
  if (N == NULL) return 0;
  return N->num;
}
int height(Node *N) {
  if (N == NULL) return 0;
  return N->height;
}
Node *newNode(int key) {
  Node *node = new Node();
  node->key = key;
  node->left = NULL;
  node->right = NULL;
  node->height = 1;
  node->num = 1;
  return node;
}
Node *rightRotate(Node *y) {
  Node *x = y->left;
  Node *T2 = x->right;
  Node *T3 = y->right;
  int tmp = y->num;
  x->num = tmp;
  y->num = 1 + get_num(T2) + get_num(T3);
  x->right = y;
  y->left = T2;
  y->height = max(height(y->left), height(y->right)) + 1;
  x->height = max(height(x->left), height(x->right)) + 1;
  return x;
}
Node *leftRotate(Node *x) {
  Node *y = x->right;
  Node *T2 = y->left;
  Node *T1 = x->left;
  int tmp = x->num;
  y->num = tmp;
  x->num = 1 + get_num(T1) + get_num(T2);
  y->left = x;
  x->right = T2;
  x->height = max(height(x->left), height(x->right)) + 1;
  y->height = max(height(y->left), height(y->right)) + 1;
  return y;
}
int getBalance(Node *N) {
  if (N == NULL) return 0;
  return height(N->left) - height(N->right);
}
Node *insert(Node *node, int key) {
  if (node == NULL) return (newNode(key));
  if (key < node->key) {
    node->left = insert(node->left, key);
    node->num += 1;
  } else if (key > node->key) {
    node->right = insert(node->right, key);
    node->num += 1;
  } else
    return node;
  node->height = 1 + max(height(node->left), height(node->right));
  int balance = getBalance(node);
  if (balance > 1 && key < node->left->key) return rightRotate(node);
  if (balance < -1 && key > node->right->key) return leftRotate(node);
  if (balance > 1 && key > node->left->key) {
    node->left = leftRotate(node->left);
    return rightRotate(node);
  }
  if (balance < -1 && key < node->right->key) {
    node->right = rightRotate(node->right);
    return leftRotate(node);
  }
  return node;
}
void inOrder(Node *root) {
  if (root != NULL) {
    inOrder(root->left);
    cout << " " << root->key << " ";
    inOrder(root->right);
  }
}
int search(Node *N, int pos) {
  int num_left = get_num(N->left);
  if (pos <= num_left) {
    return search(N->left, pos);
  } else if (pos == num_left + 1) {
    return N->key;
  } else {
    return search(N->right, pos - 1 - num_left);
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pq.push(make_pair(a[i], -i));
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> k >> pos;
    v.push_back(make_pair(k, make_pair(pos, i)));
  }
  Node *root = NULL;
  sort(v.begin(), v.end());
  int st = 0;
  int ind;
  pair<int, int> tmp;
  for (int i = 0; i < m; i++) {
    k = v[i].first;
    pos = v[i].second.first;
    ind = v[i].second.second;
    while (st < k) {
      tmp = pq.top();
      pq.pop();
      root = insert(root, -tmp.second);
      st++;
    }
    int tmp1 = search(root, pos);
    ans[ind] = a[tmp1];
  }
  for (int i = 0; i < m; i++) cout << ans[i] << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 a[200020];
priority_queue<pair<int, int> > pq;
vector<pair<int, pair<int, int> > > v;
int k, pos;
int ans[200020];
vector<int> arr;
class Node {
 public:
  int key;
  Node *left;
  Node *right;
  int height;
  int num;
};
int get_num(Node *N) {
  if (N == NULL) return 0;
  return N->num;
}
int height(Node *N) {
  if (N == NULL) return 0;
  return N->height;
}
Node *newNode(int key) {
  Node *node = new Node();
  node->key = key;
  node->left = NULL;
  node->right = NULL;
  node->height = 1;
  node->num = 1;
  return node;
}
Node *rightRotate(Node *y) {
  Node *x = y->left;
  Node *T2 = x->right;
  Node *T3 = y->right;
  int tmp = y->num;
  x->num = tmp;
  y->num = 1 + get_num(T2) + get_num(T3);
  x->right = y;
  y->left = T2;
  y->height = max(height(y->left), height(y->right)) + 1;
  x->height = max(height(x->left), height(x->right)) + 1;
  return x;
}
Node *leftRotate(Node *x) {
  Node *y = x->right;
  Node *T2 = y->left;
  Node *T1 = x->left;
  int tmp = x->num;
  y->num = tmp;
  x->num = 1 + get_num(T1) + get_num(T2);
  y->left = x;
  x->right = T2;
  x->height = max(height(x->left), height(x->right)) + 1;
  y->height = max(height(y->left), height(y->right)) + 1;
  return y;
}
int getBalance(Node *N) {
  if (N == NULL) return 0;
  return height(N->left) - height(N->right);
}
Node *insert(Node *node, int key) {
  if (node == NULL) return (newNode(key));
  if (key < node->key) {
    node->left = insert(node->left, key);
    node->num += 1;
  } else if (key > node->key) {
    node->right = insert(node->right, key);
    node->num += 1;
  } else
    return node;
  node->height = 1 + max(height(node->left), height(node->right));
  int balance = getBalance(node);
  if (balance > 1 && key < node->left->key) return rightRotate(node);
  if (balance < -1 && key > node->right->key) return leftRotate(node);
  if (balance > 1 && key > node->left->key) {
    node->left = leftRotate(node->left);
    return rightRotate(node);
  }
  if (balance < -1 && key < node->right->key) {
    node->right = rightRotate(node->right);
    return leftRotate(node);
  }
  return node;
}
void inOrder(Node *root) {
  if (root != NULL) {
    inOrder(root->left);
    cout << " " << root->key << " ";
    inOrder(root->right);
  }
}
int search(Node *N, int pos) {
  int num_left = get_num(N->left);
  if (pos <= num_left) {
    return search(N->left, pos);
  } else if (pos == num_left + 1) {
    return N->key;
  } else {
    return search(N->right, pos - 1 - num_left);
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pq.push(make_pair(a[i], -i));
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> k >> pos;
    v.push_back(make_pair(k, make_pair(pos, i)));
  }
  Node *root = NULL;
  sort(v.begin(), v.end());
  int st = 0;
  int ind;
  pair<int, int> tmp;
  for (int i = 0; i < m; i++) {
    k = v[i].first;
    pos = v[i].second.first;
    ind = v[i].second.second;
    while (st < k) {
      tmp = pq.top();
      pq.pop();
      root = insert(root, -tmp.second);
      st++;
    }
    int tmp1 = search(root, pos);
    ans[ind] = a[tmp1];
  }
  for (int i = 0; i < m; i++) cout << ans[i] << endl;
  return 0;
}

```