#include <bits/stdc++.h>
using namespace std;
const int maxN = 2e5 + 100;
struct leafy {
  int value, size;
  int l, r;
} tree[maxN * 20 + 1];
struct Query {
  int x, pos, id;
} q[maxN + 1];
struct Node {
  int x, p;
} a[maxN + 1];
int n, m, num[maxN + 1];
int res[maxN + 1], cnt, root;
inline int read() {
  int num = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) num = (num << 3) + (num << 1) + (ch ^ 48), ch = getchar();
  return num * f;
}
inline bool cmp1(Node a, Node b) {
  if (a.x == b.x) return a.p < b.p;
  return a.x > b.x;
}
inline bool cmp2(Query a, Query b) { return a.x < b.x; }
inline int new_Node(int value, int size, int l, int r) {
  tree[++cnt] = (leafy){value, size, l, r};
  return cnt;
}
inline void update(int node) {
  if (!tree[node].l) {
    tree[node].size = 1;
    return;
  }
  tree[node].size = tree[tree[node].l].size + tree[tree[node].r].size;
  tree[node].value = tree[tree[node].r].value;
}
inline void rotate(int node) {
  if (tree[tree[node].l].size > tree[tree[node].r].size * 4)
    tree[node].r =
        new_Node(tree[tree[node].r].value,
                 tree[tree[tree[node].l].r].size + tree[tree[node].r].size,
                 tree[tree[node].l].r, tree[node].r),
    tree[node].l = tree[tree[node].l].l;
  if (tree[tree[node].r].size > tree[tree[node].l].size * 4)
    tree[node].l =
        new_Node(tree[tree[tree[node].r].l].value,
                 tree[tree[node].l].size + tree[tree[tree[node].r].l].size,
                 tree[node].l, tree[tree[node].r].l),
    tree[node].r = tree[tree[node].r].r;
  update(node);
}
inline void insert(int node, int x) {
  if (tree[node].size == 1)
    tree[node].l = new_Node(min(tree[node].value, x), 1, 0, 0),
    tree[node].r = new_Node(max(tree[node].value, x), 1, 0, 0);
  else
    insert(x > tree[tree[node].l].value ? tree[node].r : tree[node].l, x);
  update(node);
  rotate(node);
}
inline int find(int node, int x) {
  if (tree[node].size == 1)
    return tree[node].value;
  else
    return x > tree[tree[node].l].size
               ? find(tree[node].r, x - tree[tree[node].l].size)
               : find(tree[node].l, x);
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i].x = num[i] = read(), a[i].p = i;
  sort(a + 1, a + n + 1, cmp1);
  m = read();
  for (int i = 1; i <= m; i++) q[i].x = read(), q[i].pos = read(), q[i].id = i;
  sort(q + 1, q + m + 1, cmp2);
  root = new_Node(1e9, 1, 0, 0);
  int now = 1;
  for (int i = 1; i <= m; i++) {
    while (now <= n && now <= q[i].x) insert(root, a[now].p), now++;
    res[q[i].id] = num[find(root, q[i].pos)];
  }
  for (int i = 1; i <= m; i++) printf("%d\n", res[i]);
  return 0;
}

### Prompt

Please formulate 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;
const int maxN = 2e5 + 100;
struct leafy {
  int value, size;
  int l, r;
} tree[maxN * 20 + 1];
struct Query {
  int x, pos, id;
} q[maxN + 1];
struct Node {
  int x, p;
} a[maxN + 1];
int n, m, num[maxN + 1];
int res[maxN + 1], cnt, root;
inline int read() {
  int num = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) num = (num << 3) + (num << 1) + (ch ^ 48), ch = getchar();
  return num * f;
}
inline bool cmp1(Node a, Node b) {
  if (a.x == b.x) return a.p < b.p;
  return a.x > b.x;
}
inline bool cmp2(Query a, Query b) { return a.x < b.x; }
inline int new_Node(int value, int size, int l, int r) {
  tree[++cnt] = (leafy){value, size, l, r};
  return cnt;
}
inline void update(int node) {
  if (!tree[node].l) {
    tree[node].size = 1;
    return;
  }
  tree[node].size = tree[tree[node].l].size + tree[tree[node].r].size;
  tree[node].value = tree[tree[node].r].value;
}
inline void rotate(int node) {
  if (tree[tree[node].l].size > tree[tree[node].r].size * 4)
    tree[node].r =
        new_Node(tree[tree[node].r].value,
                 tree[tree[tree[node].l].r].size + tree[tree[node].r].size,
                 tree[tree[node].l].r, tree[node].r),
    tree[node].l = tree[tree[node].l].l;
  if (tree[tree[node].r].size > tree[tree[node].l].size * 4)
    tree[node].l =
        new_Node(tree[tree[tree[node].r].l].value,
                 tree[tree[node].l].size + tree[tree[tree[node].r].l].size,
                 tree[node].l, tree[tree[node].r].l),
    tree[node].r = tree[tree[node].r].r;
  update(node);
}
inline void insert(int node, int x) {
  if (tree[node].size == 1)
    tree[node].l = new_Node(min(tree[node].value, x), 1, 0, 0),
    tree[node].r = new_Node(max(tree[node].value, x), 1, 0, 0);
  else
    insert(x > tree[tree[node].l].value ? tree[node].r : tree[node].l, x);
  update(node);
  rotate(node);
}
inline int find(int node, int x) {
  if (tree[node].size == 1)
    return tree[node].value;
  else
    return x > tree[tree[node].l].size
               ? find(tree[node].r, x - tree[tree[node].l].size)
               : find(tree[node].l, x);
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i].x = num[i] = read(), a[i].p = i;
  sort(a + 1, a + n + 1, cmp1);
  m = read();
  for (int i = 1; i <= m; i++) q[i].x = read(), q[i].pos = read(), q[i].id = i;
  sort(q + 1, q + m + 1, cmp2);
  root = new_Node(1e9, 1, 0, 0);
  int now = 1;
  for (int i = 1; i <= m; i++) {
    while (now <= n && now <= q[i].x) insert(root, a[now].p), now++;
    res[q[i].id] = num[find(root, q[i].pos)];
  }
  for (int i = 1; i <= m; i++) printf("%d\n", res[i]);
  return 0;
}

```