#include <bits/stdc++.h>
using namespace std;
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (fopen("1227D"
            ".inp",
            "r")) {
    freopen(
        "1227D"
        ".inp",
        "r", stdin);
    freopen(
        "1227D"
        ".out",
        "w", stdout);
  }
}
const int N = 2e5 + 2;
int n, m, res[N], st[N * 4], c[N];
pair<int, int> a[N];
pair<pair<int, int>, int> b[N];
void readinp() {
  cin >> n;
  for (int i = int(1); i <= int(n); i++) cin >> a[i].first, a[i].second = i;
  cin >> m;
  for (int i = int(1); i <= int(m); i++)
    cin >> b[i].first.first >> b[i].first.second, b[i].second = i;
}
bool cmp(pair<int, int> t1, pair<int, int> t2) {
  if (t1.first != t2.first) return t1.first > t2.first;
  return t1.second < t2.second;
}
void upd(int id, int l, int r, int p, int v) {
  if (l == r) {
    st[id] += v;
    return;
  }
  int mid = (l + r) / 2;
  if (p >= mid + 1)
    upd(id * 2 + 1, mid + 1, r, p, v);
  else
    upd(id * 2, l, mid, p, v);
  st[id] = st[id * 2] + st[id * 2 + 1];
}
int get(int id, int l, int r, int v) {
  if (l == r) return l;
  int mid = (l + r) / 2;
  if (v > st[id * 2])
    return get(id * 2 + 1, mid + 1, r, v - st[id * 2]);
  else
    return get(id * 2, l, mid, v);
}
void solve() {
  for (int i = int(1); i <= int(n); i++) c[i] = a[i].first;
  sort(a + 1, a + 1 + n, cmp);
  sort(b + 1, b + 1 + m);
  int j = 0;
  for (int i = int(1); i <= int(m); i++) {
    for (int d = int(j + 1); d <= int(b[i].first.first); d++)
      upd(1, 1, n, a[d].second, 1);
    j = b[i].first.first;
    res[b[i].second] = c[get(1, 1, n, b[i].first.second)];
  }
  for (int i = int(1); i <= int(m); i++) cout << res[i], cout << endl;
}
int main() {
  setIO();
  int t = 1;
  while (t--) {
    readinp();
    solve();
  }
}

### Prompt

In Cpp, your task is to solve 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;
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (fopen("1227D"
            ".inp",
            "r")) {
    freopen(
        "1227D"
        ".inp",
        "r", stdin);
    freopen(
        "1227D"
        ".out",
        "w", stdout);
  }
}
const int N = 2e5 + 2;
int n, m, res[N], st[N * 4], c[N];
pair<int, int> a[N];
pair<pair<int, int>, int> b[N];
void readinp() {
  cin >> n;
  for (int i = int(1); i <= int(n); i++) cin >> a[i].first, a[i].second = i;
  cin >> m;
  for (int i = int(1); i <= int(m); i++)
    cin >> b[i].first.first >> b[i].first.second, b[i].second = i;
}
bool cmp(pair<int, int> t1, pair<int, int> t2) {
  if (t1.first != t2.first) return t1.first > t2.first;
  return t1.second < t2.second;
}
void upd(int id, int l, int r, int p, int v) {
  if (l == r) {
    st[id] += v;
    return;
  }
  int mid = (l + r) / 2;
  if (p >= mid + 1)
    upd(id * 2 + 1, mid + 1, r, p, v);
  else
    upd(id * 2, l, mid, p, v);
  st[id] = st[id * 2] + st[id * 2 + 1];
}
int get(int id, int l, int r, int v) {
  if (l == r) return l;
  int mid = (l + r) / 2;
  if (v > st[id * 2])
    return get(id * 2 + 1, mid + 1, r, v - st[id * 2]);
  else
    return get(id * 2, l, mid, v);
}
void solve() {
  for (int i = int(1); i <= int(n); i++) c[i] = a[i].first;
  sort(a + 1, a + 1 + n, cmp);
  sort(b + 1, b + 1 + m);
  int j = 0;
  for (int i = int(1); i <= int(m); i++) {
    for (int d = int(j + 1); d <= int(b[i].first.first); d++)
      upd(1, 1, n, a[d].second, 1);
    j = b[i].first.first;
    res[b[i].second] = c[get(1, 1, n, b[i].first.second)];
  }
  for (int i = int(1); i <= int(m); i++) cout << res[i], cout << endl;
}
int main() {
  setIO();
  int t = 1;
  while (t--) {
    readinp();
    solve();
  }
}

```