#include <bits/stdc++.h>
using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pil = pair<int, ll>;
using pll = pair<ll, ll>;
template <typename T>
using vec = vector<T>;
using vi = vec<int>;
using vl = vec<ll>;
template <typename T>
using que = queue<T>;
template <typename T>
using deq = deque<T>;
template <typename T>
T id(T b) {
  return b;
};
template <typename T>
void chmax(T &x, T y) {
  if (x < y) x = y;
}
template <typename T>
void chmin(T &x, T y) {
  if (x > y) x = y;
}
template <typename S, typename K>
bool contains(S &s, K k) {
  return s.find(k) != s.end();
}
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
constexpr ll TEN(int n) {
  if (n == 0)
    return 1LL;
  else
    return 10LL * TEN(n - 1);
}
class Segtree {
  using ll = long long;
  int n;
  vector<ll> dat;
  const ll empty = 0;
  inline ll concat(ll a, ll b) { return a + b; }
  ll query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return empty;
    if (a <= l && r <= b)
      return dat[k];
    else
      return concat(query(a, b, 2 * k + 1, l, (l + r) / 2),
                    query(a, b, 2 * k + 2, (l + r) / 2, r));
  }

 public:
  Segtree() = default;
  Segtree(int n_) {
    n = 1;
    while (n < n_) n *= 2;
    dat.assign(2 * n - 1, 0);
  }
  void update(int k, ll a) {
    k += n - 1;
    dat[k] = a;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = concat(dat[2 * k + 1], dat[2 * k + 2]);
    }
  }
  ll query(int a, int b) { return query(a, b, 0, 0, n); }
};
const int MAX_N = 200000;
int n, m;
pii a[MAX_N];
int b[MAX_N];
vec<pii> query[MAX_N + 1];
int main() {
  fastio();
  cin >> n;
  for (int i = 0; i < int(n); i++) {
    cin >> a[i].first;
    b[i] = a[i].first;
    a[i].second = i;
  }
  sort(a, a + n, [](pii a, pii b) {
    return a.first > b.first or (a.first == b.first and a.second < b.second);
  });
  cin >> m;
  for (int i = 0; i < int(m); i++) {
    int k, pos;
    cin >> k >> pos;
    query[k].emplace_back(i, pos);
  }
  vi ans(m);
  Segtree st(n);
  auto check = [&](int m, int pos) { return st.query(0, m + 1) >= pos; };
  for (int i = 1; i <= int(n); i++) {
    st.update(a[i - 1].second, 1);
    for (auto [j, pos] : query[i]) {
      int l = -1, r = n - 1;
      while (r - l > 1) {
        int mid = (l + r) / 2;
        if (check(mid, pos)) {
          r = mid;
        } else {
          l = mid;
        }
      }
      ans[j] = b[r];
    }
  }
  for (auto x : ans) {
    cout << x << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pil = pair<int, ll>;
using pll = pair<ll, ll>;
template <typename T>
using vec = vector<T>;
using vi = vec<int>;
using vl = vec<ll>;
template <typename T>
using que = queue<T>;
template <typename T>
using deq = deque<T>;
template <typename T>
T id(T b) {
  return b;
};
template <typename T>
void chmax(T &x, T y) {
  if (x < y) x = y;
}
template <typename T>
void chmin(T &x, T y) {
  if (x > y) x = y;
}
template <typename S, typename K>
bool contains(S &s, K k) {
  return s.find(k) != s.end();
}
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
constexpr ll TEN(int n) {
  if (n == 0)
    return 1LL;
  else
    return 10LL * TEN(n - 1);
}
class Segtree {
  using ll = long long;
  int n;
  vector<ll> dat;
  const ll empty = 0;
  inline ll concat(ll a, ll b) { return a + b; }
  ll query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return empty;
    if (a <= l && r <= b)
      return dat[k];
    else
      return concat(query(a, b, 2 * k + 1, l, (l + r) / 2),
                    query(a, b, 2 * k + 2, (l + r) / 2, r));
  }

 public:
  Segtree() = default;
  Segtree(int n_) {
    n = 1;
    while (n < n_) n *= 2;
    dat.assign(2 * n - 1, 0);
  }
  void update(int k, ll a) {
    k += n - 1;
    dat[k] = a;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = concat(dat[2 * k + 1], dat[2 * k + 2]);
    }
  }
  ll query(int a, int b) { return query(a, b, 0, 0, n); }
};
const int MAX_N = 200000;
int n, m;
pii a[MAX_N];
int b[MAX_N];
vec<pii> query[MAX_N + 1];
int main() {
  fastio();
  cin >> n;
  for (int i = 0; i < int(n); i++) {
    cin >> a[i].first;
    b[i] = a[i].first;
    a[i].second = i;
  }
  sort(a, a + n, [](pii a, pii b) {
    return a.first > b.first or (a.first == b.first and a.second < b.second);
  });
  cin >> m;
  for (int i = 0; i < int(m); i++) {
    int k, pos;
    cin >> k >> pos;
    query[k].emplace_back(i, pos);
  }
  vi ans(m);
  Segtree st(n);
  auto check = [&](int m, int pos) { return st.query(0, m + 1) >= pos; };
  for (int i = 1; i <= int(n); i++) {
    st.update(a[i - 1].second, 1);
    for (auto [j, pos] : query[i]) {
      int l = -1, r = n - 1;
      while (r - l > 1) {
        int mid = (l + r) / 2;
        if (check(mid, pos)) {
          r = mid;
        } else {
          l = mid;
        }
      }
      ans[j] = b[r];
    }
  }
  for (auto x : ans) {
    cout << x << "\n";
  }
  return 0;
}

```