#include <bits/stdc++.h>
using namespace std;
int tree[4 * 200001 + 1];
int A[200001 + 1], B[200001 + 1];
int op(int x, int y) {
  if (A[x] == A[y])
    return min(x, y);
  else
    return A[x] > A[y] ? x : y;
}
void build(int node, int start, int end) {
  if (start == end)
    tree[node] = start;
  else {
    int mid = (start + end) / 2;
    build(2 * node, start, mid);
    build(2 * node + 1, mid + 1, end);
    tree[node] = op(tree[2 * node], tree[2 * node + 1]);
  }
}
void update(int node, int start, int end, int idx) {
  if (start == end)
    return;
  else {
    int mid = (start + end) / 2;
    if (start <= idx && idx <= mid)
      update(2 * node, start, mid, idx);
    else
      update(2 * node + 1, mid + 1, end, idx);
    tree[node] = op(tree[2 * node], tree[2 * node + 1]);
  }
}
int query(int node, int start, int end, int l, int r) {
  if (r < start || end < l) return 0;
  if (l <= start && end <= r) return tree[node];
  int mid = (start + end) / 2;
  int p1 = query(2 * node, start, mid, l, r);
  int p2 = query(2 * node + 1, mid + 1, end, l, r);
  return op(p1, p2);
}
int N;
int BIT[200001];
void updateBIT(int i) {
  while (i <= N) {
    BIT[i]++;
    i += (i & -i);
  }
}
int sumBIT(int i) {
  int s = 0;
  while (i > 0) {
    s += BIT[i];
    i -= (i & -i);
  }
  return s;
}
int indice(int x) {
  int lo = 1, hi = N, mid, ind;
  while (lo <= hi) {
    mid = (lo + hi) / 2;
    if (sumBIT(mid) >= x) {
      hi = mid - 1;
      ind = mid;
    } else {
      lo = mid + 1;
    }
  }
  return ind;
}
vector<pair<int, int> > queries;
vector<int> ordem;
int ans[200001];
int cmp(int a, int b) { return queries[a] < queries[b]; }
int main() {
  int M, sz = 0, inicial;
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &A[i]);
    B[i] = A[i];
  }
  inicial = N;
  build(1, 1, N);
  scanf("%d", &M);
  for (int i = 0; i < M; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    queries.push_back({x, y});
    ordem.push_back(i);
  }
  sort(ordem.begin(), ordem.end(), cmp);
  for (int i = 0; i < M; i++) {
    while (sz != queries[ordem[i]].first) {
      updateBIT(tree[1]);
      A[tree[1]] = 0;
      update(1, 1, N, tree[1]);
      sz++;
    }
    ans[ordem[i]] = B[indice(queries[ordem[i]].second)];
  }
  for (int i = 0; i < M; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
int tree[4 * 200001 + 1];
int A[200001 + 1], B[200001 + 1];
int op(int x, int y) {
  if (A[x] == A[y])
    return min(x, y);
  else
    return A[x] > A[y] ? x : y;
}
void build(int node, int start, int end) {
  if (start == end)
    tree[node] = start;
  else {
    int mid = (start + end) / 2;
    build(2 * node, start, mid);
    build(2 * node + 1, mid + 1, end);
    tree[node] = op(tree[2 * node], tree[2 * node + 1]);
  }
}
void update(int node, int start, int end, int idx) {
  if (start == end)
    return;
  else {
    int mid = (start + end) / 2;
    if (start <= idx && idx <= mid)
      update(2 * node, start, mid, idx);
    else
      update(2 * node + 1, mid + 1, end, idx);
    tree[node] = op(tree[2 * node], tree[2 * node + 1]);
  }
}
int query(int node, int start, int end, int l, int r) {
  if (r < start || end < l) return 0;
  if (l <= start && end <= r) return tree[node];
  int mid = (start + end) / 2;
  int p1 = query(2 * node, start, mid, l, r);
  int p2 = query(2 * node + 1, mid + 1, end, l, r);
  return op(p1, p2);
}
int N;
int BIT[200001];
void updateBIT(int i) {
  while (i <= N) {
    BIT[i]++;
    i += (i & -i);
  }
}
int sumBIT(int i) {
  int s = 0;
  while (i > 0) {
    s += BIT[i];
    i -= (i & -i);
  }
  return s;
}
int indice(int x) {
  int lo = 1, hi = N, mid, ind;
  while (lo <= hi) {
    mid = (lo + hi) / 2;
    if (sumBIT(mid) >= x) {
      hi = mid - 1;
      ind = mid;
    } else {
      lo = mid + 1;
    }
  }
  return ind;
}
vector<pair<int, int> > queries;
vector<int> ordem;
int ans[200001];
int cmp(int a, int b) { return queries[a] < queries[b]; }
int main() {
  int M, sz = 0, inicial;
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &A[i]);
    B[i] = A[i];
  }
  inicial = N;
  build(1, 1, N);
  scanf("%d", &M);
  for (int i = 0; i < M; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    queries.push_back({x, y});
    ordem.push_back(i);
  }
  sort(ordem.begin(), ordem.end(), cmp);
  for (int i = 0; i < M; i++) {
    while (sz != queries[ordem[i]].first) {
      updateBIT(tree[1]);
      A[tree[1]] = 0;
      update(1, 1, N, tree[1]);
      sz++;
    }
    ans[ordem[i]] = B[indice(queries[ordem[i]].second)];
  }
  for (int i = 0; i < M; i++) printf("%d\n", ans[i]);
  return 0;
}

```