#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 200000;
const int MAXQ = 200000;
const int MAXS = 4 * MAXN;
int n;
int a[MAXN];
int nq;
int qk[MAXQ], qidx[MAXQ], qans[MAXQ];
pair<int, int> o[MAXN];
vector<int> byk[MAXN + 1];
int sval[MAXS];
void sinit(int x, int l, int r) {
  if (l == r) {
    sval[x] = 0;
  } else {
    int m = l + (r - l) / 2;
    sinit(2 * x + 1, l, m);
    sinit(2 * x + 2, m + 1, r);
    sval[x] = sval[2 * x + 1] + sval[2 * x + 2];
  }
}
void smod(int x, int l, int r, int IDX) {
  if (l == r) {
    ++sval[x];
  } else {
    int m = l + (r - l) / 2;
    if (IDX <= m)
      smod(2 * x + 1, l, m, IDX);
    else
      smod(2 * x + 2, m + 1, r, IDX);
    sval[x] = sval[2 * x + 1] + sval[2 * x + 2];
  }
}
int sget(int x, int l, int r, int OFF) {
  assert(OFF < sval[x]);
  if (l == r) {
    return l;
  } else {
    int m = l + (r - l) / 2;
    if (OFF < sval[2 * x + 1])
      return sget(2 * x + 1, l, m, OFF);
    else
      return sget(2 * x + 2, m + 1, r, OFF - sval[2 * x + 1]);
  }
}
void solve() {
  for (int i = (0); i < (n); ++i) o[i] = make_pair(-a[i], i);
  sort(o, o + n);
  for (int i = (0); i <= (n); ++i) byk[i].clear();
  for (int i = (0); i < (nq); ++i) byk[qk[i]].push_back(i);
  sinit(0, 0, n - 1);
  for (int i = (0); i < (n); ++i) {
    smod(0, 0, n - 1, o[i].second);
    for (int j = (0); j < (((int)(byk[i + 1]).size())); ++j) {
      int id = byk[i + 1][j];
      int idx = sget(0, 0, n - 1, qidx[id]);
      qans[id] = a[idx];
    }
  }
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  scanf("%d", &nq);
  for (int i = (0); i < (nq); ++i) scanf("%d%d", &qk[i], &qidx[i]), --qidx[i];
  solve();
  for (int i = (0); i < (nq); ++i) printf("%d\n", qans[i]);
}
int main() {
  run();
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 200000;
const int MAXQ = 200000;
const int MAXS = 4 * MAXN;
int n;
int a[MAXN];
int nq;
int qk[MAXQ], qidx[MAXQ], qans[MAXQ];
pair<int, int> o[MAXN];
vector<int> byk[MAXN + 1];
int sval[MAXS];
void sinit(int x, int l, int r) {
  if (l == r) {
    sval[x] = 0;
  } else {
    int m = l + (r - l) / 2;
    sinit(2 * x + 1, l, m);
    sinit(2 * x + 2, m + 1, r);
    sval[x] = sval[2 * x + 1] + sval[2 * x + 2];
  }
}
void smod(int x, int l, int r, int IDX) {
  if (l == r) {
    ++sval[x];
  } else {
    int m = l + (r - l) / 2;
    if (IDX <= m)
      smod(2 * x + 1, l, m, IDX);
    else
      smod(2 * x + 2, m + 1, r, IDX);
    sval[x] = sval[2 * x + 1] + sval[2 * x + 2];
  }
}
int sget(int x, int l, int r, int OFF) {
  assert(OFF < sval[x]);
  if (l == r) {
    return l;
  } else {
    int m = l + (r - l) / 2;
    if (OFF < sval[2 * x + 1])
      return sget(2 * x + 1, l, m, OFF);
    else
      return sget(2 * x + 2, m + 1, r, OFF - sval[2 * x + 1]);
  }
}
void solve() {
  for (int i = (0); i < (n); ++i) o[i] = make_pair(-a[i], i);
  sort(o, o + n);
  for (int i = (0); i <= (n); ++i) byk[i].clear();
  for (int i = (0); i < (nq); ++i) byk[qk[i]].push_back(i);
  sinit(0, 0, n - 1);
  for (int i = (0); i < (n); ++i) {
    smod(0, 0, n - 1, o[i].second);
    for (int j = (0); j < (((int)(byk[i + 1]).size())); ++j) {
      int id = byk[i + 1][j];
      int idx = sget(0, 0, n - 1, qidx[id]);
      qans[id] = a[idx];
    }
  }
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  scanf("%d", &nq);
  for (int i = (0); i < (nq); ++i) scanf("%d%d", &qk[i], &qidx[i]), --qidx[i];
  solve();
  for (int i = (0); i < (nq); ++i) printf("%d\n", qans[i]);
}
int main() {
  run();
  return 0;
}

```