#include <bits/stdc++.h>
using namespace std;
class SegmentTree {
 public:
  vector<vector<long long> > a;
  long long n;
  SegmentTree(vector<long long>& arr) {
    n = arr.size();
    a.resize(4 * n);
    build(1, 0, arr.size(), arr);
  }
  void build(long long v, long long vl, long long vr, vector<long long>& arr) {
    if (vr - vl == 1) {
      a[v].push_back(arr[vl]);
    } else {
      long long vm = (vl + vr) / 2;
      build(v * 2, vl, vm, arr);
      build(v * 2 + 1, vm, vr, arr);
      a[v].resize(vr - vl);
      merge(a[v * 2].begin(), a[v * 2].end(), a[v * 2 + 1].begin(),
            a[v * 2 + 1].end(), a[v].begin());
    }
  }
  long long get(long long v, long long vl, long long vr, long long l,
                long long r, long long x) {
    if (vl == l && vr == r) {
      return a[v].end() - lower_bound(a[v].begin(), a[v].end(), x);
    } else {
      long long vm = (vl + vr) / 2;
      if (r <= vm) {
        return get(v * 2, vl, vm, l, r, x);
      } else if (l >= vm) {
        return get(v * 2 + 1, vm, vr, l, r, x);
      } else {
        return get(v * 2, vl, vm, l, vm, x) + get(v * 2 + 1, vm, vr, vm, r, x);
      }
    }
  }
  long long get(long long l, long long r, long long x) {
    return get(1, 0, n, l, r, x);
  }
};
signed main() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> nums = a;
  sort(nums.begin(), nums.end());
  nums.erase(unique(nums.begin(), nums.end()), nums.end());
  vector<vector<long long> > posof(nums.size());
  for (long long i = 0; i < n; i++) {
    posof[lower_bound(nums.begin(), nums.end(), a[i]) - nums.begin()].push_back(
        i);
  }
  vector<long long> sufcntof(posof.size());
  sufcntof.back() = posof.back().size();
  for (long long i = (long long)posof.size() - 2; i >= 0; i--) {
    sufcntof[i] = sufcntof[i + 1] + posof[i].size();
  }
  SegmentTree st(a);
  long long m;
  cin >> m;
  for (long long i = 0; i < m; i++) {
    long long k, pos;
    cin >> k >> pos;
    long long numpos = sufcntof.rend() -
                       lower_bound(sufcntof.rbegin(), sufcntof.rend(), k) - 1;
    long long num = nums[numpos];
    long long limit =
        k - (numpos == sufcntof.size() - 1 ? 0 : sufcntof[numpos + 1]);
    long long l = 0;
    long long r = n;
    while (r - l > 1) {
      long long m = (l + r) / 2;
      long long c = st.get(0, m + 1, num + 1);
      c += min((long long)(lower_bound(posof[numpos].begin(),
                                       posof[numpos].end(), m + 1) -
                           posof[numpos].begin()),
               limit);
      if (c >= pos) {
        r = m;
      } else {
        l = m;
      }
    }
    long long m = l;
    long long c = st.get(0, m + 1, num + 1);
    c += min((long long)(lower_bound(posof[numpos].begin(), posof[numpos].end(),
                                     m + 1) -
                         posof[numpos].begin()),
             limit);
    if (c >= pos) {
      r = m;
    }
    cout << a[r] << "\n";
  }
  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;
class SegmentTree {
 public:
  vector<vector<long long> > a;
  long long n;
  SegmentTree(vector<long long>& arr) {
    n = arr.size();
    a.resize(4 * n);
    build(1, 0, arr.size(), arr);
  }
  void build(long long v, long long vl, long long vr, vector<long long>& arr) {
    if (vr - vl == 1) {
      a[v].push_back(arr[vl]);
    } else {
      long long vm = (vl + vr) / 2;
      build(v * 2, vl, vm, arr);
      build(v * 2 + 1, vm, vr, arr);
      a[v].resize(vr - vl);
      merge(a[v * 2].begin(), a[v * 2].end(), a[v * 2 + 1].begin(),
            a[v * 2 + 1].end(), a[v].begin());
    }
  }
  long long get(long long v, long long vl, long long vr, long long l,
                long long r, long long x) {
    if (vl == l && vr == r) {
      return a[v].end() - lower_bound(a[v].begin(), a[v].end(), x);
    } else {
      long long vm = (vl + vr) / 2;
      if (r <= vm) {
        return get(v * 2, vl, vm, l, r, x);
      } else if (l >= vm) {
        return get(v * 2 + 1, vm, vr, l, r, x);
      } else {
        return get(v * 2, vl, vm, l, vm, x) + get(v * 2 + 1, vm, vr, vm, r, x);
      }
    }
  }
  long long get(long long l, long long r, long long x) {
    return get(1, 0, n, l, r, x);
  }
};
signed main() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> nums = a;
  sort(nums.begin(), nums.end());
  nums.erase(unique(nums.begin(), nums.end()), nums.end());
  vector<vector<long long> > posof(nums.size());
  for (long long i = 0; i < n; i++) {
    posof[lower_bound(nums.begin(), nums.end(), a[i]) - nums.begin()].push_back(
        i);
  }
  vector<long long> sufcntof(posof.size());
  sufcntof.back() = posof.back().size();
  for (long long i = (long long)posof.size() - 2; i >= 0; i--) {
    sufcntof[i] = sufcntof[i + 1] + posof[i].size();
  }
  SegmentTree st(a);
  long long m;
  cin >> m;
  for (long long i = 0; i < m; i++) {
    long long k, pos;
    cin >> k >> pos;
    long long numpos = sufcntof.rend() -
                       lower_bound(sufcntof.rbegin(), sufcntof.rend(), k) - 1;
    long long num = nums[numpos];
    long long limit =
        k - (numpos == sufcntof.size() - 1 ? 0 : sufcntof[numpos + 1]);
    long long l = 0;
    long long r = n;
    while (r - l > 1) {
      long long m = (l + r) / 2;
      long long c = st.get(0, m + 1, num + 1);
      c += min((long long)(lower_bound(posof[numpos].begin(),
                                       posof[numpos].end(), m + 1) -
                           posof[numpos].begin()),
               limit);
      if (c >= pos) {
        r = m;
      } else {
        l = m;
      }
    }
    long long m = l;
    long long c = st.get(0, m + 1, num + 1);
    c += min((long long)(lower_bound(posof[numpos].begin(), posof[numpos].end(),
                                     m + 1) -
                         posof[numpos].begin()),
             limit);
    if (c >= pos) {
      r = m;
    }
    cout << a[r] << "\n";
  }
  return 0;
}

```