#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 10;
long long n, Q, gt[N], res[N];
struct num {
  long long v;
  long long p;
};
struct query {
  long long k;
  long long pos;
  long long p;
};
bool cmp(num A, num B) {
  if (A.v == B.v) {
    return A.p < B.p;
  }
  return A.v > B.v;
}
bool cmpq(query A, query B) { return A.k < B.k; }
num a[N];
query q[N];
long long st[4 * N];
void Update(long long id, long long l, long long r, long long x, long long v) {
  if (x < l || r < x) {
    return;
  }
  if (l == r) {
    st[id] = v;
    return;
  }
  long long mid = (l + r) / 2;
  Update(id * 2, l, mid, x, v);
  Update(id * 2 + 1, mid + 1, r, x, v);
  st[id] = st[id * 2] + st[id * 2 + 1];
}
long long Find(long long id, long long l, long long r, long long v) {
  if (st[id] < v) {
    return -1;
  }
  if (l == r) {
    return l;
  }
  long long mid = (l + r) / 2;
  long long res = Find(id * 2, l, mid, v);
  if (res == -1) {
    res = Find(id * 2 + 1, mid + 1, r, v - st[id * 2]);
  }
  return res;
}
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i].v;
    gt[i] = a[i].v;
    a[i].p = i;
  }
  sort(a + 1, a + n + 1, cmp);
  cin >> Q;
  for (long long i = 1; i <= Q; i++) {
    cin >> q[i].k >> q[i].pos;
    q[i].p = i;
  }
  sort(q + 1, q + Q + 1, cmpq);
  q[0].k = 0;
  for (long long i = 1; i <= Q; i++) {
    long long k = q[i].k;
    long long pos = q[i].pos;
    long long p = q[i].p;
    for (long long j = q[i - 1].k + 1; j <= k; j++) {
      Update(1, 1, n, a[j].p, 1);
    }
    long long vt = Find(1, 1, n, pos);
    res[p] = gt[vt];
  }
  for (long long i = 1; i <= Q; i++) {
    cout << res[i] << "\n";
  }
}

### Prompt

Generate 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;
const long long N = 2e5 + 10;
long long n, Q, gt[N], res[N];
struct num {
  long long v;
  long long p;
};
struct query {
  long long k;
  long long pos;
  long long p;
};
bool cmp(num A, num B) {
  if (A.v == B.v) {
    return A.p < B.p;
  }
  return A.v > B.v;
}
bool cmpq(query A, query B) { return A.k < B.k; }
num a[N];
query q[N];
long long st[4 * N];
void Update(long long id, long long l, long long r, long long x, long long v) {
  if (x < l || r < x) {
    return;
  }
  if (l == r) {
    st[id] = v;
    return;
  }
  long long mid = (l + r) / 2;
  Update(id * 2, l, mid, x, v);
  Update(id * 2 + 1, mid + 1, r, x, v);
  st[id] = st[id * 2] + st[id * 2 + 1];
}
long long Find(long long id, long long l, long long r, long long v) {
  if (st[id] < v) {
    return -1;
  }
  if (l == r) {
    return l;
  }
  long long mid = (l + r) / 2;
  long long res = Find(id * 2, l, mid, v);
  if (res == -1) {
    res = Find(id * 2 + 1, mid + 1, r, v - st[id * 2]);
  }
  return res;
}
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i].v;
    gt[i] = a[i].v;
    a[i].p = i;
  }
  sort(a + 1, a + n + 1, cmp);
  cin >> Q;
  for (long long i = 1; i <= Q; i++) {
    cin >> q[i].k >> q[i].pos;
    q[i].p = i;
  }
  sort(q + 1, q + Q + 1, cmpq);
  q[0].k = 0;
  for (long long i = 1; i <= Q; i++) {
    long long k = q[i].k;
    long long pos = q[i].pos;
    long long p = q[i].p;
    for (long long j = q[i - 1].k + 1; j <= k; j++) {
      Update(1, 1, n, a[j].p, 1);
    }
    long long vt = Find(1, 1, n, pos);
    res[p] = gt[vt];
  }
  for (long long i = 1; i <= Q; i++) {
    cout << res[i] << "\n";
  }
}

```