#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int out = 0, fh = 1;
  char jp = getchar();
  while ((jp > '9' || jp < '0') && jp != '-') jp = getchar();
  if (jp == '-') fh = -1, jp = getchar();
  while (jp >= '0' && jp <= '9') out = out * 10 + jp - '0', jp = getchar();
  return out * fh;
}
const int MAXN = 2e5 + 10;
int n, m, k, pos, a[MAXN];
struct info {
  int val, pos;
  bool operator<(const info &rhs) const {
    if (val != rhs.val) return val > rhs.val;
    return pos < rhs.pos;
  }
} p[MAXN];
struct node {
  int ls, rs, siz;
  node() { ls = rs = siz = 0; }
} Tree[MAXN * 20];
int idx = 0, rt[MAXN];
void upd(int &o, int pre, int l, int r, int pos) {
  o = ++idx;
  Tree[o] = Tree[pre];
  ++Tree[o].siz;
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (pos <= mid)
    upd(Tree[o].ls, Tree[pre].ls, l, mid, pos);
  else
    upd(Tree[o].rs, Tree[pre].rs, mid + 1, r, pos);
}
int query(int o, int l, int r, int k) {
  if (l == r) return a[l];
  int mid = (l + r) >> 1;
  if (Tree[Tree[o].ls].siz >= k)
    return query(Tree[o].ls, l, mid, k);
  else
    return query(Tree[o].rs, mid + 1, r, k - Tree[Tree[o].ls].siz);
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
    p[i].val = a[i];
    p[i].pos = i;
  }
  sort(p + 1, p + 1 + n);
  for (int i = 1; i <= n; ++i) upd(rt[i], rt[i - 1], 1, n, p[i].pos);
  m = read();
  for (int i = 1; i <= m; ++i) {
    k = read(), pos = read();
    printf("%d\n", query(rt[k], 1, n, pos));
  }
  return 0;
}

### Prompt

Generate 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;
inline int read() {
  int out = 0, fh = 1;
  char jp = getchar();
  while ((jp > '9' || jp < '0') && jp != '-') jp = getchar();
  if (jp == '-') fh = -1, jp = getchar();
  while (jp >= '0' && jp <= '9') out = out * 10 + jp - '0', jp = getchar();
  return out * fh;
}
const int MAXN = 2e5 + 10;
int n, m, k, pos, a[MAXN];
struct info {
  int val, pos;
  bool operator<(const info &rhs) const {
    if (val != rhs.val) return val > rhs.val;
    return pos < rhs.pos;
  }
} p[MAXN];
struct node {
  int ls, rs, siz;
  node() { ls = rs = siz = 0; }
} Tree[MAXN * 20];
int idx = 0, rt[MAXN];
void upd(int &o, int pre, int l, int r, int pos) {
  o = ++idx;
  Tree[o] = Tree[pre];
  ++Tree[o].siz;
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (pos <= mid)
    upd(Tree[o].ls, Tree[pre].ls, l, mid, pos);
  else
    upd(Tree[o].rs, Tree[pre].rs, mid + 1, r, pos);
}
int query(int o, int l, int r, int k) {
  if (l == r) return a[l];
  int mid = (l + r) >> 1;
  if (Tree[Tree[o].ls].siz >= k)
    return query(Tree[o].ls, l, mid, k);
  else
    return query(Tree[o].rs, mid + 1, r, k - Tree[Tree[o].ls].siz);
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
    p[i].val = a[i];
    p[i].pos = i;
  }
  sort(p + 1, p + 1 + n);
  for (int i = 1; i <= n; ++i) upd(rt[i], rt[i - 1], 1, n, p[i].pos);
  m = read();
  for (int i = 1; i <= m; ++i) {
    k = read(), pos = read();
    printf("%d\n", query(rt[k], 1, n, pos));
  }
  return 0;
}

```