#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("03")
#pragma GCC optimization("unroll-loops")
using namespace std;
const int maxn = 200001;
const int maxe = 20;
int a[maxn], b[maxn], c, k, pa[maxn], x, y, q, sol, d, d2, mod = 1000000007, j;
int m, n;
char h;
set<int> se;
string s;
pair<int, int> p[maxn + 1];
int rmq[maxn][maxe];
int st[4 * maxn], la[4 * maxn], ft[maxn], dt[maxn], dt2[maxn];
void build(int n) {
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 1; i + (1 << j) - 1 <= n; i++) {
      rmq[i][j] = min(rmq[i][j - 1], rmq[i + (1 << (j - 1))][j - 1]);
    }
  }
}
int query(int a, int b) {
  int k = 31 - __builtin_clz(b - a + 1);
  return min(rmq[a][k], rmq[b - (1 << k) + 1][k]);
}
int fin(int i) {
  if (pa[i] == i) return i;
  return pa[i] = fin(pa[i]);
}
int bs(int l, int r) {
  while (l != r) {
    int mi = (l + r) / 2;
    if (a[mi] == x) return mi;
    if (a[mi] > x)
      r = mi;
    else
      l = mi + 1;
  }
  return l;
}
void crear(int l, int r, int nod) {
  if (l == r)
    st[nod] = a[l];
  else {
    int mi = (l + r) / 2;
    crear(l, mi, 2 * nod);
    crear(mi + 1, r, 2 * nod + 1);
    st[nod] = min(st[2 * nod], st[2 * nod + 1]);
  }
}
void upd(int l, int r, int nod) {
  if (l > x || r < x) return;
  if (la[nod] != 0) {
    st[nod] += la[nod];
    la[2 * nod] += la[nod];
    la[2 * nod + 1] += la[nod];
    la[nod] = 0;
  }
  if (l == r)
    st[nod] += y;
  else {
    int mi = (l + r) / 2;
    upd(l, mi, 2 * nod);
    upd(mi + 1, r, 2 * nod + 1);
    st[nod] = min(st[2 * nod], st[2 * nod + 1]);
  }
}
int qu(int l, int r, int nod) {
  if (l > y || r < x) return 100001;
  if (la[nod] != 0) {
    st[nod] += la[nod];
    la[2 * nod] += la[nod];
    la[2 * nod + 1] += la[nod];
    la[nod] = 0;
  }
  if (l >= x && r <= y)
    return st[nod];
  else {
    int mi = (l + r) / 2;
    return min(qu(l, mi, 2 * nod), qu(mi + 1, r, 2 * nod + 1));
  }
}
void lazy(int l, int r, int nod) {
  if (l > y || r < x) return;
  if (la[nod] != 0) {
    st[nod] += la[nod];
    la[2 * nod] += la[nod];
    la[2 * nod + 1] += la[nod];
    la[nod] = 0;
  }
  if (l >= x && r <= y) {
    st[nod] += q;
    if (l != r) {
      la[nod * 2] += q;
      la[2 * nod + 1] += q;
    }
  } else {
    int mi = (l + r) / 2;
    lazy(l, mi, 2 * nod);
    lazy(mi + 1, r, 2 * nod + 1);
    st[nod] = min(st[2 * nod], st[2 * nod + 1]);
  }
}
int ftq(int n) {
  int c = 0;
  while (n) {
    c += ft[n];
    n -= (n & -n);
  }
  return c;
}
void ftupd(int x) {
  for (int i = x; i <= n; i += (i & -i)) ft[i] += y;
}
void abi(int x) {
  for (int i = x; i <= n; i += (i & -i)) dt[i] += y;
}
void abi2(int x) {
  for (int i = x; i <= n; i += (i & -i)) dt2[i] += y;
}
int q1(int n) {
  int c = 0;
  while (n) {
    c += dt[n];
    n -= (n & -n);
  }
  return c;
}
int q2(int n) {
  int c = 0;
  while (n) {
    c += dt[n];
    n -= (n & -n);
  }
  return c;
}
void ini() {
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    y = a[i] - a[i - 1];
    abi(i);
    y = a[i] - a[i - 1] * i;
    abi2(i);
  }
}
int main() {
  ios_base::sync_with_stdio();
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  cin >> m;
  while (m--) {
    priority_queue<pair<int, int>> p, sol;
    for (int i = 1; i <= n; i++) {
      p.push({a[i], -i});
    }
    cin >> x >> y;
    for (int i = 1; i <= x; i++) {
      sol.push({p.top().second, p.top().first});
      p.pop();
    }
    for (int i = 1; i < y; i++) {
      sol.pop();
    }
    cout << sol.top().second << '\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>
#pragma GCC target("avx2")
#pragma GCC optimization("03")
#pragma GCC optimization("unroll-loops")
using namespace std;
const int maxn = 200001;
const int maxe = 20;
int a[maxn], b[maxn], c, k, pa[maxn], x, y, q, sol, d, d2, mod = 1000000007, j;
int m, n;
char h;
set<int> se;
string s;
pair<int, int> p[maxn + 1];
int rmq[maxn][maxe];
int st[4 * maxn], la[4 * maxn], ft[maxn], dt[maxn], dt2[maxn];
void build(int n) {
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 1; i + (1 << j) - 1 <= n; i++) {
      rmq[i][j] = min(rmq[i][j - 1], rmq[i + (1 << (j - 1))][j - 1]);
    }
  }
}
int query(int a, int b) {
  int k = 31 - __builtin_clz(b - a + 1);
  return min(rmq[a][k], rmq[b - (1 << k) + 1][k]);
}
int fin(int i) {
  if (pa[i] == i) return i;
  return pa[i] = fin(pa[i]);
}
int bs(int l, int r) {
  while (l != r) {
    int mi = (l + r) / 2;
    if (a[mi] == x) return mi;
    if (a[mi] > x)
      r = mi;
    else
      l = mi + 1;
  }
  return l;
}
void crear(int l, int r, int nod) {
  if (l == r)
    st[nod] = a[l];
  else {
    int mi = (l + r) / 2;
    crear(l, mi, 2 * nod);
    crear(mi + 1, r, 2 * nod + 1);
    st[nod] = min(st[2 * nod], st[2 * nod + 1]);
  }
}
void upd(int l, int r, int nod) {
  if (l > x || r < x) return;
  if (la[nod] != 0) {
    st[nod] += la[nod];
    la[2 * nod] += la[nod];
    la[2 * nod + 1] += la[nod];
    la[nod] = 0;
  }
  if (l == r)
    st[nod] += y;
  else {
    int mi = (l + r) / 2;
    upd(l, mi, 2 * nod);
    upd(mi + 1, r, 2 * nod + 1);
    st[nod] = min(st[2 * nod], st[2 * nod + 1]);
  }
}
int qu(int l, int r, int nod) {
  if (l > y || r < x) return 100001;
  if (la[nod] != 0) {
    st[nod] += la[nod];
    la[2 * nod] += la[nod];
    la[2 * nod + 1] += la[nod];
    la[nod] = 0;
  }
  if (l >= x && r <= y)
    return st[nod];
  else {
    int mi = (l + r) / 2;
    return min(qu(l, mi, 2 * nod), qu(mi + 1, r, 2 * nod + 1));
  }
}
void lazy(int l, int r, int nod) {
  if (l > y || r < x) return;
  if (la[nod] != 0) {
    st[nod] += la[nod];
    la[2 * nod] += la[nod];
    la[2 * nod + 1] += la[nod];
    la[nod] = 0;
  }
  if (l >= x && r <= y) {
    st[nod] += q;
    if (l != r) {
      la[nod * 2] += q;
      la[2 * nod + 1] += q;
    }
  } else {
    int mi = (l + r) / 2;
    lazy(l, mi, 2 * nod);
    lazy(mi + 1, r, 2 * nod + 1);
    st[nod] = min(st[2 * nod], st[2 * nod + 1]);
  }
}
int ftq(int n) {
  int c = 0;
  while (n) {
    c += ft[n];
    n -= (n & -n);
  }
  return c;
}
void ftupd(int x) {
  for (int i = x; i <= n; i += (i & -i)) ft[i] += y;
}
void abi(int x) {
  for (int i = x; i <= n; i += (i & -i)) dt[i] += y;
}
void abi2(int x) {
  for (int i = x; i <= n; i += (i & -i)) dt2[i] += y;
}
int q1(int n) {
  int c = 0;
  while (n) {
    c += dt[n];
    n -= (n & -n);
  }
  return c;
}
int q2(int n) {
  int c = 0;
  while (n) {
    c += dt[n];
    n -= (n & -n);
  }
  return c;
}
void ini() {
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    y = a[i] - a[i - 1];
    abi(i);
    y = a[i] - a[i - 1] * i;
    abi2(i);
  }
}
int main() {
  ios_base::sync_with_stdio();
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  cin >> m;
  while (m--) {
    priority_queue<pair<int, int>> p, sol;
    for (int i = 1; i <= n; i++) {
      p.push({a[i], -i});
    }
    cin >> x >> y;
    for (int i = 1; i <= x; i++) {
      sol.push({p.top().second, p.top().first});
      p.pop();
    }
    for (int i = 1; i < y; i++) {
      sol.pop();
    }
    cout << sol.top().second << '\n';
  }
}

```