#include <bits/stdc++.h>
using namespace std;
namespace io {
const int SI = 1 << 21 | 1;
char IB[SI], *IS, *IT, OB[SI], *OS = OB, *OT = OS + SI - 1, c, ch[100];
int f, t;
inline void flush() { fwrite(OB, 1, OS - OB, stdout), OS = OB; }
inline void pc(char x) {
  *OS++ = x;
  if (OS == OT) flush();
}
template <class I>
inline void rd(I &x) {
  for (f = 1, c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                              IS == IT ? EOF : *IS++)
                            : *IS++);
       c < '0' || c > '9';
       c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++))
    if (c == '-') f = -1;
  for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + (c & 15),
      c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                      IS == IT ? EOF : *IS++)
                    : *IS++))
    ;
  x *= f;
}
inline void rds(char *s, int &x) {
  for (c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++);
       c < 33 || c > 126;
       c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++))
    ;
  for (x = 0; c >= 33 && c <= 126;
       s[++x] = c, c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                                   IS == IT ? EOF : *IS++)
                                 : *IS++))
    ;
  s[x + 1] = '\0';
}
template <class I>
inline void print(I x, char k = '\n') {
  if (!x) pc('0');
  if (x < 0) pc('-'), x = -x;
  while (x) ch[++t] = x % 10 + '0', x /= 10;
  while (t) pc(ch[t--]);
  pc(k);
}
inline void prints(string s) {
  int x = s.length();
  while (t < x) pc(s[t++]);
  pc('\n'), t = 0;
}
struct Flush {
  ~Flush() { flush(); }
} flusher;
}  // namespace io
using io::print;
using io::prints;
using io::rd;
using io::rds;
const int N = 2e5 + 7;
int n, m, a[N], c[N], ans[N];
priority_queue<pair<int, int> > q;
pair<pair<int, int>, int> p[N];
inline void add(int x) {
  while (x <= n) c[x]++, x += x & -x;
}
inline int ask(int x) {
  int k = 0;
  while (x) k += c[x], x -= x & -x;
  return k;
}
inline int qry(int x) {
  int l = 1, r = n;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (ask(mid) < x)
      l = mid + 1;
    else
      r = mid;
  }
  return a[l];
}
int main() {
  rd(n);
  for (int i = 1; i <= n; i++) {
    rd(a[i]);
    q.push(make_pair(a[i], -i));
  }
  rd(m);
  for (int i = 1, x, y; i <= m; i++) {
    rd(x), rd(y), p[i] = make_pair(make_pair(x, y), i);
  }
  sort(p + 1, p + m + 1);
  int t = 0;
  for (int i = 1; i <= m; i++) {
    while (t < p[i].first.first) {
      int x = -q.top().second;
      q.pop();
      add(x);
      ++t;
    }
    ans[p[i].second] = qry(p[i].first.second);
  }
  for (int i = 1; i <= m; i++) print(ans[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;
namespace io {
const int SI = 1 << 21 | 1;
char IB[SI], *IS, *IT, OB[SI], *OS = OB, *OT = OS + SI - 1, c, ch[100];
int f, t;
inline void flush() { fwrite(OB, 1, OS - OB, stdout), OS = OB; }
inline void pc(char x) {
  *OS++ = x;
  if (OS == OT) flush();
}
template <class I>
inline void rd(I &x) {
  for (f = 1, c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                              IS == IT ? EOF : *IS++)
                            : *IS++);
       c < '0' || c > '9';
       c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++))
    if (c == '-') f = -1;
  for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + (c & 15),
      c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                      IS == IT ? EOF : *IS++)
                    : *IS++))
    ;
  x *= f;
}
inline void rds(char *s, int &x) {
  for (c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++);
       c < 33 || c > 126;
       c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++))
    ;
  for (x = 0; c >= 33 && c <= 126;
       s[++x] = c, c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                                   IS == IT ? EOF : *IS++)
                                 : *IS++))
    ;
  s[x + 1] = '\0';
}
template <class I>
inline void print(I x, char k = '\n') {
  if (!x) pc('0');
  if (x < 0) pc('-'), x = -x;
  while (x) ch[++t] = x % 10 + '0', x /= 10;
  while (t) pc(ch[t--]);
  pc(k);
}
inline void prints(string s) {
  int x = s.length();
  while (t < x) pc(s[t++]);
  pc('\n'), t = 0;
}
struct Flush {
  ~Flush() { flush(); }
} flusher;
}  // namespace io
using io::print;
using io::prints;
using io::rd;
using io::rds;
const int N = 2e5 + 7;
int n, m, a[N], c[N], ans[N];
priority_queue<pair<int, int> > q;
pair<pair<int, int>, int> p[N];
inline void add(int x) {
  while (x <= n) c[x]++, x += x & -x;
}
inline int ask(int x) {
  int k = 0;
  while (x) k += c[x], x -= x & -x;
  return k;
}
inline int qry(int x) {
  int l = 1, r = n;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (ask(mid) < x)
      l = mid + 1;
    else
      r = mid;
  }
  return a[l];
}
int main() {
  rd(n);
  for (int i = 1; i <= n; i++) {
    rd(a[i]);
    q.push(make_pair(a[i], -i));
  }
  rd(m);
  for (int i = 1, x, y; i <= m; i++) {
    rd(x), rd(y), p[i] = make_pair(make_pair(x, y), i);
  }
  sort(p + 1, p + m + 1);
  int t = 0;
  for (int i = 1; i <= m; i++) {
    while (t < p[i].first.first) {
      int x = -q.top().second;
      q.pop();
      add(x);
      ++t;
    }
    ans[p[i].second] = qry(p[i].first.second);
  }
  for (int i = 1; i <= m; i++) print(ans[i]);
  return 0;
}

```