#include <bits/stdc++.h>
using namespace std;
inline void print(int n) {
  if (n == 0) {
    putchar('0');
    putchar('\n');
  } else if (n == -1) {
    putchar('-');
    putchar('1');
    putchar('\n');
  } else {
    char buf[11];
    buf[10] = '\n';
    int i = 9;
    while (n) {
      buf[i--] = n % 10 + '0';
      n /= 10;
    }
    while (buf[i] != '\n') putchar(buf[++i]);
  }
}
int read() {
  int cc = getc(stdin);
  for (; cc < '0' || cc > '9';) cc = getc(stdin);
  int ret = 0;
  for (; cc >= '0' && cc <= '9';) {
    ret = ret * 10 + cc - '0';
    cc = getc(stdin);
  }
  return ret;
}
long long power(long long num, long long g) {
  if (g == 0) return 1;
  if (g % 2 == 1) return (num * power((num * num), g / 2));
  return power((num * num), g / 2);
}
int tree[800005];
void built(int start, int end, int index) {
  if (start == end) {
    tree[index] = 0;
    return;
  }
  int mid = (start + end) / 2;
  built(start, mid, 2 * index);
  built(mid + 1, end, 1 + 2 * index);
}
void update(int start, int end, int index, int ind) {
  if (start == end) {
    tree[index] = 1;
    return;
  }
  int mid = (start + end) / 2;
  if (ind <= mid)
    update(start, mid, 2 * index, ind);
  else
    update(mid + 1, end, 2 * index + 1, ind);
  tree[index] = tree[2 * index] + tree[2 * index + 1];
}
int search(int start, int end, int index, int l, int r) {
  if (start > r || end < l) return 0;
  if (start >= l && end <= r) return tree[index];
  int mid = (start + end) / 2;
  return search(start, mid, 2 * index, l, r) +
         search(mid + 1, end, 2 * index + 1, l, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int a[n + 1];
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      pq;
  for (int i = 0; i < n; i++) {
    cin >> a[i + 1];
    pq.push(make_pair(a[i + 1], -i - 1));
  }
  int q;
  cin >> q;
  pair<pair<int, int>, int> pa[q];
  for (int i = 0; i < q; i++) {
    cin >> pa[i].first.first >> pa[i].first.second;
    pa[i].second = i;
  }
  int ans[q];
  sort(pa, pa + q);
  int z = q - 1;
  int hd = n;
  while (z >= 0) {
    while (z >= 0 && pa[z].first.first == hd) {
      int start = 1, end = n;
      while (start < end) {
        int mid = (start + end) / 2;
        int sd = search(1, n, 1, 1, mid);
        if (mid - sd >= pa[z].first.second)
          end = mid;
        else
          start = mid + 1;
      }
      ans[pa[z].second] = a[start];
      z--;
    }
    int sd = -pq.top().second;
    update(1, n, 1, sd);
    pq.pop();
    hd--;
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << endl;
  }
  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;
inline void print(int n) {
  if (n == 0) {
    putchar('0');
    putchar('\n');
  } else if (n == -1) {
    putchar('-');
    putchar('1');
    putchar('\n');
  } else {
    char buf[11];
    buf[10] = '\n';
    int i = 9;
    while (n) {
      buf[i--] = n % 10 + '0';
      n /= 10;
    }
    while (buf[i] != '\n') putchar(buf[++i]);
  }
}
int read() {
  int cc = getc(stdin);
  for (; cc < '0' || cc > '9';) cc = getc(stdin);
  int ret = 0;
  for (; cc >= '0' && cc <= '9';) {
    ret = ret * 10 + cc - '0';
    cc = getc(stdin);
  }
  return ret;
}
long long power(long long num, long long g) {
  if (g == 0) return 1;
  if (g % 2 == 1) return (num * power((num * num), g / 2));
  return power((num * num), g / 2);
}
int tree[800005];
void built(int start, int end, int index) {
  if (start == end) {
    tree[index] = 0;
    return;
  }
  int mid = (start + end) / 2;
  built(start, mid, 2 * index);
  built(mid + 1, end, 1 + 2 * index);
}
void update(int start, int end, int index, int ind) {
  if (start == end) {
    tree[index] = 1;
    return;
  }
  int mid = (start + end) / 2;
  if (ind <= mid)
    update(start, mid, 2 * index, ind);
  else
    update(mid + 1, end, 2 * index + 1, ind);
  tree[index] = tree[2 * index] + tree[2 * index + 1];
}
int search(int start, int end, int index, int l, int r) {
  if (start > r || end < l) return 0;
  if (start >= l && end <= r) return tree[index];
  int mid = (start + end) / 2;
  return search(start, mid, 2 * index, l, r) +
         search(mid + 1, end, 2 * index + 1, l, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int a[n + 1];
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      pq;
  for (int i = 0; i < n; i++) {
    cin >> a[i + 1];
    pq.push(make_pair(a[i + 1], -i - 1));
  }
  int q;
  cin >> q;
  pair<pair<int, int>, int> pa[q];
  for (int i = 0; i < q; i++) {
    cin >> pa[i].first.first >> pa[i].first.second;
    pa[i].second = i;
  }
  int ans[q];
  sort(pa, pa + q);
  int z = q - 1;
  int hd = n;
  while (z >= 0) {
    while (z >= 0 && pa[z].first.first == hd) {
      int start = 1, end = n;
      while (start < end) {
        int mid = (start + end) / 2;
        int sd = search(1, n, 1, 1, mid);
        if (mid - sd >= pa[z].first.second)
          end = mid;
        else
          start = mid + 1;
      }
      ans[pa[z].second] = a[start];
      z--;
    }
    int sd = -pq.top().second;
    update(1, n, 1, sd);
    pq.pop();
    hd--;
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}

```