#include <bits/stdc++.h>
using namespace std;
long long N = 200000;
vector<bool> visited(N + 1, false);
bool isprime(long long x) {
  for (long long i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) return false;
  }
  return true;
}
void findfact(long long x, map<long long, vector<long long>>& m) {
  for (long long i = 3; i * i <= x; i += 2) {
    if (x % i == 0) {
      if (i == x / i)
        m[x].push_back(i);
      else {
        m[x].push_back(i);
        m[x].push_back(x / i);
      }
      break;
    }
  }
  return;
}
struct cmp {
  bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
    if (a.first > b.first)
      return false;
    else if (a.first < b.first)
      return true;
    if (a.second < b.second) return false;
    return true;
  }
};
bool isPalindrome(string t) {
  long long st = 0;
  long long end = t.size() - 1;
  while (st <= end) {
    if (t[st] != t[end]) return false;
    st++;
    end--;
  }
  return true;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool pow2(long long x) { return x && (!(x & (x - 1))); }
unsigned long long factorial(unsigned long long n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
unsigned long long binomialCoeff(unsigned long long n, unsigned long long k) {
  unsigned long long res = 1;
  if (k > n - k) k = n - k;
  for (unsigned long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
vector<long long> hp;
void primeFactors(long long n) {
  while (n % 2 == 0) {
    hp.push_back(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      hp.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) hp.push_back(n);
}
long long minFlipsMonoIncr(string S) {
  int flip = 0;
  int ones = 0;
  bool flag = false;
  for (int i = 0; i < S.size(); i++) {
    if (S[i] - '0' == 1) flag = true;
    if (flag && S[i] - '0' == 0)
      flip++;
    else if (S[i] == '1')
      ones++;
    if (flip > ones) flip = ones;
  }
  return flip;
}
void dfs() {}
int main() {
  ios_base::sync_with_stdio(0);
  long long n;
  cin >> n;
  vector<long long> arr(n + 1);
  for (long long i = 1; i <= n; i++) cin >> arr[i];
  long long m;
  cin >> m;
  while (m--) {
    long long k, pos;
    cin >> k >> pos;
    priority_queue<pair<long long, long long>,
                   vector<pair<long long, long long>>, cmp>
        pq;
    for (long long i = 1; i <= n; i++) pq.push(make_pair(arr[i], i));
    vector<long long> temp(n + 1, -1);
    while (k--) {
      temp[pq.top().second] = pq.top().first;
      pq.pop();
    }
    long long ele = 0;
    long long i;
    for (i = 1; i <= n; i++) {
      if (temp[i] == -1)
        continue;
      else
        ele++;
      if (ele == pos) break;
    }
    cout << temp[i] << "\n";
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 N = 200000;
vector<bool> visited(N + 1, false);
bool isprime(long long x) {
  for (long long i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) return false;
  }
  return true;
}
void findfact(long long x, map<long long, vector<long long>>& m) {
  for (long long i = 3; i * i <= x; i += 2) {
    if (x % i == 0) {
      if (i == x / i)
        m[x].push_back(i);
      else {
        m[x].push_back(i);
        m[x].push_back(x / i);
      }
      break;
    }
  }
  return;
}
struct cmp {
  bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
    if (a.first > b.first)
      return false;
    else if (a.first < b.first)
      return true;
    if (a.second < b.second) return false;
    return true;
  }
};
bool isPalindrome(string t) {
  long long st = 0;
  long long end = t.size() - 1;
  while (st <= end) {
    if (t[st] != t[end]) return false;
    st++;
    end--;
  }
  return true;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool pow2(long long x) { return x && (!(x & (x - 1))); }
unsigned long long factorial(unsigned long long n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
unsigned long long binomialCoeff(unsigned long long n, unsigned long long k) {
  unsigned long long res = 1;
  if (k > n - k) k = n - k;
  for (unsigned long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
vector<long long> hp;
void primeFactors(long long n) {
  while (n % 2 == 0) {
    hp.push_back(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      hp.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) hp.push_back(n);
}
long long minFlipsMonoIncr(string S) {
  int flip = 0;
  int ones = 0;
  bool flag = false;
  for (int i = 0; i < S.size(); i++) {
    if (S[i] - '0' == 1) flag = true;
    if (flag && S[i] - '0' == 0)
      flip++;
    else if (S[i] == '1')
      ones++;
    if (flip > ones) flip = ones;
  }
  return flip;
}
void dfs() {}
int main() {
  ios_base::sync_with_stdio(0);
  long long n;
  cin >> n;
  vector<long long> arr(n + 1);
  for (long long i = 1; i <= n; i++) cin >> arr[i];
  long long m;
  cin >> m;
  while (m--) {
    long long k, pos;
    cin >> k >> pos;
    priority_queue<pair<long long, long long>,
                   vector<pair<long long, long long>>, cmp>
        pq;
    for (long long i = 1; i <= n; i++) pq.push(make_pair(arr[i], i));
    vector<long long> temp(n + 1, -1);
    while (k--) {
      temp[pq.top().second] = pq.top().first;
      pq.pop();
    }
    long long ele = 0;
    long long i;
    for (i = 1; i <= n; i++) {
      if (temp[i] == -1)
        continue;
      else
        ele++;
      if (ele == pos) break;
    }
    cout << temp[i] << "\n";
  }
}

```