#include <bits/stdc++.h>
using namespace std;
int read() {
  int w = 0, f = 1;
  char c = ' ';
  while (c < '0' || c > '9') c = getchar(), f = c == '-' ? -1 : f;
  while (c >= '0' && c <= '9') w = w * 10 + c - 48, c = getchar();
  return w * f;
}
int A[200005], ch[200005][2], val[200005], pri[200005], siz[200005], sz;
void update(int x) { siz[x] = 1 + siz[ch[x][0]] + siz[ch[x][1]]; }
int ne(int v) {
  siz[++sz] = 1;
  val[sz] = v;
  pri[sz] = rand();
  return sz;
}
int merge(int x, int y) {
  if (!x || !y) return x + y;
  if (pri[x] < pri[y]) {
    ch[x][1] = merge(ch[x][1], y);
    update(x);
    return x;
  } else {
    ch[y][0] = merge(x, ch[y][0]);
    update(y);
    return y;
  }
}
void split(int now, int k, int &x, int &y) {
  if (!now)
    x = y = 0;
  else {
    if (val[now] <= k)
      x = now, split(ch[now][1], k, ch[now][1], y);
    else
      y = now, split(ch[now][0], k, x, ch[now][0]);
    update(now);
  }
}
int kth(int now, int k) {
  while (1) {
    if (k <= siz[ch[now][0]])
      now = ch[now][0];
    else if (k == siz[ch[now][0]] + 1)
      return now;
    else
      k -= siz[ch[now][0]] + 1, now = ch[now][1];
  }
}
struct node1 {
  int val, pos;
} a[200005];
struct node2 {
  int k, p, pos, ans;
} q[200005];
bool cmp1(node1 x, node1 y) {
  return x.val != y.val ? x.val > y.val : x.pos < y.pos;
}
bool cmp2(node2 x, node2 y) { return x.k < y.k; }
bool cmp3(node2 x, node2 y) { return x.pos < y.pos; }
int n, m;
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) A[i] = a[i].val = read(), a[i].pos = i;
  sort(a + 1, a + n + 1, cmp1);
  m = read();
  for (int i = 1; i <= m; i++) q[i].k = read(), q[i].p = read(), q[i].pos = i;
  sort(q + 1, q + m + 1, cmp2);
  int nowk = 0, root = 0, x, y;
  for (int i = 1; i <= m; i++) {
    while (nowk != q[i].k) {
      nowk++;
      split(root, a[nowk].pos, x, y);
      root = merge(merge(x, ne(a[nowk].pos)), y);
    }
    q[i].ans = A[val[kth(root, q[i].p)]];
  }
  sort(q + 1, q + m + 1, cmp3);
  for (int i = 1; i <= m; i++) printf("%d\n", q[i].ans);
  return 0;
}

### Prompt

Your task is to create 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;
int read() {
  int w = 0, f = 1;
  char c = ' ';
  while (c < '0' || c > '9') c = getchar(), f = c == '-' ? -1 : f;
  while (c >= '0' && c <= '9') w = w * 10 + c - 48, c = getchar();
  return w * f;
}
int A[200005], ch[200005][2], val[200005], pri[200005], siz[200005], sz;
void update(int x) { siz[x] = 1 + siz[ch[x][0]] + siz[ch[x][1]]; }
int ne(int v) {
  siz[++sz] = 1;
  val[sz] = v;
  pri[sz] = rand();
  return sz;
}
int merge(int x, int y) {
  if (!x || !y) return x + y;
  if (pri[x] < pri[y]) {
    ch[x][1] = merge(ch[x][1], y);
    update(x);
    return x;
  } else {
    ch[y][0] = merge(x, ch[y][0]);
    update(y);
    return y;
  }
}
void split(int now, int k, int &x, int &y) {
  if (!now)
    x = y = 0;
  else {
    if (val[now] <= k)
      x = now, split(ch[now][1], k, ch[now][1], y);
    else
      y = now, split(ch[now][0], k, x, ch[now][0]);
    update(now);
  }
}
int kth(int now, int k) {
  while (1) {
    if (k <= siz[ch[now][0]])
      now = ch[now][0];
    else if (k == siz[ch[now][0]] + 1)
      return now;
    else
      k -= siz[ch[now][0]] + 1, now = ch[now][1];
  }
}
struct node1 {
  int val, pos;
} a[200005];
struct node2 {
  int k, p, pos, ans;
} q[200005];
bool cmp1(node1 x, node1 y) {
  return x.val != y.val ? x.val > y.val : x.pos < y.pos;
}
bool cmp2(node2 x, node2 y) { return x.k < y.k; }
bool cmp3(node2 x, node2 y) { return x.pos < y.pos; }
int n, m;
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) A[i] = a[i].val = read(), a[i].pos = i;
  sort(a + 1, a + n + 1, cmp1);
  m = read();
  for (int i = 1; i <= m; i++) q[i].k = read(), q[i].p = read(), q[i].pos = i;
  sort(q + 1, q + m + 1, cmp2);
  int nowk = 0, root = 0, x, y;
  for (int i = 1; i <= m; i++) {
    while (nowk != q[i].k) {
      nowk++;
      split(root, a[nowk].pos, x, y);
      root = merge(merge(x, ne(a[nowk].pos)), y);
    }
    q[i].ans = A[val[kth(root, q[i].p)]];
  }
  sort(q + 1, q + m + 1, cmp3);
  for (int i = 1; i <= m; i++) printf("%d\n", q[i].ans);
  return 0;
}

```