#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 200010;
const int INF = 2147483600;
int N, M;
struct data {
  int d, id, pos;
} a[MAXN + 1], q[MAXN + 1];
int A[MAXN + 1], Ans[MAXN + 1];
bool cmpd(data a, data b) {
  if (a.d != b.d) return a.d > b.d;
  return a.pos < b.pos;
}
bool cmpk(data a, data b) { return a.d < b.d; }
inline int lowbit(int x) { return x & (-x); }
int c[MAXN + 1];
inline int Query(int x) {
  int ans = 0;
  for (; x; x -= lowbit(x)) {
    ans += c[x];
  }
  return ans;
}
inline void Add(int k, int x) {
  for (; k <= N; k += lowbit(k)) c[k] += x;
  return;
}
int main() {
  N = read();
  for (int i = 1; i <= N; i++) {
    a[i].pos = i;
    a[i].d = read();
    A[i] = a[i].d;
  }
  sort(a + 1, a + N + 1, cmpd);
  M = read();
  for (int i = 1; i <= M; i++) {
    q[i].d = read(), q[i].pos = read();
    q[i].id = i;
  }
  sort(q + 1, q + M + 1, cmpk);
  int nw = 0;
  for (int i = 1; i <= N; i++) {
    Add(a[i].pos, 1);
    while (nw < M && q[nw + 1].d == i) {
      int l = 1, r = N, ans = 0;
      while (l <= r) {
        int mid = (l + r) >> 1;
        if (Query(mid) < q[nw + 1].pos)
          ans = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      Ans[q[nw + 1].id] = ans + 1;
      ++nw;
    }
  }
  for (int i = 1; i <= M; i++) printf("%d\n", A[Ans[i]]);
  return 0;
}

### Prompt

Create a solution in CPP for 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 x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 200010;
const int INF = 2147483600;
int N, M;
struct data {
  int d, id, pos;
} a[MAXN + 1], q[MAXN + 1];
int A[MAXN + 1], Ans[MAXN + 1];
bool cmpd(data a, data b) {
  if (a.d != b.d) return a.d > b.d;
  return a.pos < b.pos;
}
bool cmpk(data a, data b) { return a.d < b.d; }
inline int lowbit(int x) { return x & (-x); }
int c[MAXN + 1];
inline int Query(int x) {
  int ans = 0;
  for (; x; x -= lowbit(x)) {
    ans += c[x];
  }
  return ans;
}
inline void Add(int k, int x) {
  for (; k <= N; k += lowbit(k)) c[k] += x;
  return;
}
int main() {
  N = read();
  for (int i = 1; i <= N; i++) {
    a[i].pos = i;
    a[i].d = read();
    A[i] = a[i].d;
  }
  sort(a + 1, a + N + 1, cmpd);
  M = read();
  for (int i = 1; i <= M; i++) {
    q[i].d = read(), q[i].pos = read();
    q[i].id = i;
  }
  sort(q + 1, q + M + 1, cmpk);
  int nw = 0;
  for (int i = 1; i <= N; i++) {
    Add(a[i].pos, 1);
    while (nw < M && q[nw + 1].d == i) {
      int l = 1, r = N, ans = 0;
      while (l <= r) {
        int mid = (l + r) >> 1;
        if (Query(mid) < q[nw + 1].pos)
          ans = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      Ans[q[nw + 1].id] = ans + 1;
      ++nw;
    }
  }
  for (int i = 1; i <= M; i++) printf("%d\n", A[Ans[i]]);
  return 0;
}

```