#include <bits/stdc++.h>
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
unsigned long long power(unsigned long long a, unsigned long long b) {
  a %= 1000000007;
  unsigned long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return res % 1000000007;
}
long long modmul(long long x, long long y) { return (x * y) % 1000000007; }
long long modadd(long long x, long long y) { return (x + y) % 1000000007; }
long long modsub(long long x, long long y) {
  return (x - y + 1000000007) % 1000000007;
}
long long fact[1000007] = {0};
void facto() {
  fact[0] = 1;
  fact[1] = 1;
  for (long long i = 2; i < 100007; i++)
    fact[i] = (fact[i - 1] * i) % 1000000007;
}
long long ncr(long long n, long long r) {
  long long res = 1;
  res = fact[n];
  res = (res * (power(fact[r], 1000000007 - 2))) % 1000000007;
  res = (res * (power(fact[n - r], 1000000007 - 2))) % 1000000007;
  return res;
}
long long npr(long long n, long long r) {
  long long res = 1;
  res = fact[n];
  res = (res * (power(fact[n - r], 1000000007 - 2))) % 1000000007;
  return res;
}
inline long long toint(const std::string &s) {
  std::stringstream ss;
  ss << s;
  long long x;
  ss >> x;
  return x;
}
inline std::string tostring(long long number) {
  std::stringstream ss;
  ss << number;
  return ss.str();
}
inline std::string tobin(long long x) { return std::bitset<63>(x).to_string(); }
using namespace std;
long long a[101] = {0};
bool vis[101] = {0};
long long solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i < n + 1; i++) cin >> a[i];
  long long t;
  cin >> t;
  while (t--) {
    long long k, pos;
    cin >> k >> pos;
    memset(vis, false, sizeof(vis));
    while (k--) {
      long long p = 0;
      for (long long i = 1; i <= n; i++) {
        if (!vis[i] && a[i] > a[p]) p = i;
      }
      vis[p] = true;
    }
    long long ans = 0;
    while (pos--) {
      while (!vis[++ans])
        ;
    }
    cout << a[ans] << '\n';
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
}

### 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>
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
unsigned long long power(unsigned long long a, unsigned long long b) {
  a %= 1000000007;
  unsigned long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return res % 1000000007;
}
long long modmul(long long x, long long y) { return (x * y) % 1000000007; }
long long modadd(long long x, long long y) { return (x + y) % 1000000007; }
long long modsub(long long x, long long y) {
  return (x - y + 1000000007) % 1000000007;
}
long long fact[1000007] = {0};
void facto() {
  fact[0] = 1;
  fact[1] = 1;
  for (long long i = 2; i < 100007; i++)
    fact[i] = (fact[i - 1] * i) % 1000000007;
}
long long ncr(long long n, long long r) {
  long long res = 1;
  res = fact[n];
  res = (res * (power(fact[r], 1000000007 - 2))) % 1000000007;
  res = (res * (power(fact[n - r], 1000000007 - 2))) % 1000000007;
  return res;
}
long long npr(long long n, long long r) {
  long long res = 1;
  res = fact[n];
  res = (res * (power(fact[n - r], 1000000007 - 2))) % 1000000007;
  return res;
}
inline long long toint(const std::string &s) {
  std::stringstream ss;
  ss << s;
  long long x;
  ss >> x;
  return x;
}
inline std::string tostring(long long number) {
  std::stringstream ss;
  ss << number;
  return ss.str();
}
inline std::string tobin(long long x) { return std::bitset<63>(x).to_string(); }
using namespace std;
long long a[101] = {0};
bool vis[101] = {0};
long long solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i < n + 1; i++) cin >> a[i];
  long long t;
  cin >> t;
  while (t--) {
    long long k, pos;
    cin >> k >> pos;
    memset(vis, false, sizeof(vis));
    while (k--) {
      long long p = 0;
      for (long long i = 1; i <= n; i++) {
        if (!vis[i] && a[i] > a[p]) p = i;
      }
      vis[p] = true;
    }
    long long ans = 0;
    while (pos--) {
      while (!vis[++ans])
        ;
    }
    cout << a[ans] << '\n';
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
}

```