#include <bits/stdc++.h>
using namespace std;
int n;
int b[200010];
pair<int, int> a[200010];
pair<int, int> c[200010];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first > b.first) return true;
  if (a.first < b.first) return false;
  return a.second < b.second;
}
struct node {
  int val;
  node *left, *right;
  node() {}
  node(node* l, node* r, int v) {
    left = l;
    right = r;
    val = v;
  }
};
void build(node* n, int l, int r) {
  if (l == r) {
    n->val = 0;
    return;
  }
  n->left = new node(NULL, NULL, 0);
  n->right = new node(NULL, NULL, 0);
  build(n->left, l, (l + r) / 2);
  build(n->right, (l + r) / 2 + 1, r);
}
void update(node* pre, node* cur, int l, int r, int u, int v) {
  if (r < u || l > u) return;
  if (l == r) {
    cur->val = v;
    return;
  }
  int mid = (l + r) / 2;
  if (u <= mid) {
    cur->right = pre->right;
    cur->left = new node(NULL, NULL, 0);
    update(pre->left, cur->left, l, mid, u, v);
  } else {
    cur->left = pre->left;
    cur->right = new node(NULL, NULL, 0);
    update(pre->right, cur->right, mid + 1, r, u, v);
  }
  cur->val = cur->left->val + cur->right->val;
}
int get(node* a, int l, int r, int u, int v) {
  if (u > r || v < l) return 0;
  if (u <= l && r <= v) {
    return a->val;
  }
  int mid = (l + r) / 2;
  return get(a->left, l, mid, u, v) + get(a->right, mid + 1, r, u, v);
}
node* version[200010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    b[i] = x;
    a[i] = pair<int, int>(x, i);
  }
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    c[i] = pair<int, int>(a[i].second, i);
  }
  sort(c + 1, c + n + 1);
  node* root = new node(NULL, NULL, 0);
  build(root, 1, n);
  version[0] = root;
  for (int i = 1; i <= n; i++) {
    version[i] = new node(NULL, NULL, 0);
    update(version[i - 1], version[i], 1, n, c[i].second, 1);
  }
  int Q;
  cin >> Q;
  for (int i = 1; i <= Q; i++) {
    int k, pos;
    cin >> k >> pos;
    int l = 0, r = n + 1, mid;
    while (r - l > 1) {
      mid = (l + r) / 2;
      int temp = get(version[mid], 1, n, 1, k);
      if (temp >= pos)
        r = mid;
      else
        l = mid;
    }
    int ind = c[r].second;
    cout << a[ind].first << "\n";
  }
}

### 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;
int n;
int b[200010];
pair<int, int> a[200010];
pair<int, int> c[200010];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first > b.first) return true;
  if (a.first < b.first) return false;
  return a.second < b.second;
}
struct node {
  int val;
  node *left, *right;
  node() {}
  node(node* l, node* r, int v) {
    left = l;
    right = r;
    val = v;
  }
};
void build(node* n, int l, int r) {
  if (l == r) {
    n->val = 0;
    return;
  }
  n->left = new node(NULL, NULL, 0);
  n->right = new node(NULL, NULL, 0);
  build(n->left, l, (l + r) / 2);
  build(n->right, (l + r) / 2 + 1, r);
}
void update(node* pre, node* cur, int l, int r, int u, int v) {
  if (r < u || l > u) return;
  if (l == r) {
    cur->val = v;
    return;
  }
  int mid = (l + r) / 2;
  if (u <= mid) {
    cur->right = pre->right;
    cur->left = new node(NULL, NULL, 0);
    update(pre->left, cur->left, l, mid, u, v);
  } else {
    cur->left = pre->left;
    cur->right = new node(NULL, NULL, 0);
    update(pre->right, cur->right, mid + 1, r, u, v);
  }
  cur->val = cur->left->val + cur->right->val;
}
int get(node* a, int l, int r, int u, int v) {
  if (u > r || v < l) return 0;
  if (u <= l && r <= v) {
    return a->val;
  }
  int mid = (l + r) / 2;
  return get(a->left, l, mid, u, v) + get(a->right, mid + 1, r, u, v);
}
node* version[200010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    b[i] = x;
    a[i] = pair<int, int>(x, i);
  }
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    c[i] = pair<int, int>(a[i].second, i);
  }
  sort(c + 1, c + n + 1);
  node* root = new node(NULL, NULL, 0);
  build(root, 1, n);
  version[0] = root;
  for (int i = 1; i <= n; i++) {
    version[i] = new node(NULL, NULL, 0);
    update(version[i - 1], version[i], 1, n, c[i].second, 1);
  }
  int Q;
  cin >> Q;
  for (int i = 1; i <= Q; i++) {
    int k, pos;
    cin >> k >> pos;
    int l = 0, r = n + 1, mid;
    while (r - l > 1) {
      mid = (l + r) / 2;
      int temp = get(version[mid], 1, n, 1, k);
      if (temp >= pos)
        r = mid;
      else
        l = mid;
    }
    int ind = c[r].second;
    cout << a[ind].first << "\n";
  }
}

```