#include <bits/stdc++.h>
using namespace std;
struct node {
  int val, pos, sz, pr;
  node *l, *r;
};
node *t;
int ans[200020];
pair<int, int> b[200020];
pair<pair<int, int>, int> q[200020];
bool cmp(pair<int, int> x, pair<int, int> y) {
  return (x.first > y.first) || (x.first == y.first && x.second < y.second);
}
bool cmp2(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  return (x.first.first < y.first.first) ||
         (x.first.first == y.first.first && x.first.second < y.first.second);
}
int get_size(node *t) { return (t == 0) ? 0 : t->sz; }
void Upd(node *t) {
  if (t != 0) t->sz = get_size(t->l) + get_size(t->r) + 1;
}
node *Merge(node *tl, node *tr) {
  if (tl == 0) return tr;
  if (tr == 0) return tl;
  if (tl->pr < tr->pr) {
    tr->l = Merge(tl, tr->l);
    Upd(tr);
    return tr;
  } else {
    tl->r = Merge(tl->r, tr);
    Upd(tl);
    return tl;
  }
}
void Split(node *t, node *&tl, node *&tr, int x) {
  if (t == 0) {
    tl = tr = 0;
    return;
  }
  if (t->pos < x) {
    Split(t->r, t->r, tr, x);
    tl = t;
    Upd(t);
  } else {
    Split(t->l, tl, t->l, x);
    tr = t;
    Upd(t);
  }
}
void Split2(node *t, node *&tl, node *&tr, int x) {
  if (t == 0) {
    tl = tr = 0;
    return;
  }
  if (get_size(t->l) + 1 < x) {
    int newx = x - get_size(t->l) - 1;
    Split2(t->r, t->r, tr, newx);
    tl = t;
    Upd(t);
  } else {
    Split2(t->l, tl, t->l, x);
    tr = t;
    Upd(t);
  }
}
void Out(node *t) {
  if (t == 0) return;
  Out(t->l);
  cout << t->val << " ";
  Out(t->r);
}
void Insert(node *&t, pair<int, int> x) {
  node *tl, *tmr, *tm, *tr;
  Split(t, tl, tmr, x.second);
  Split(tmr, tm, tr, x.second + 1);
  tm = new node;
  tm->pos = x.second;
  tm->pr = rand() * rand();
  tm->sz = 1;
  tm->val = x.first;
  tm->l = tm->r = 0;
  t = Merge(tl, Merge(tm, tr));
}
int Find(node *&t, int x) {
  node *tlm, *tr, *tl, *tm;
  Split2(t, tlm, tr, x + 1);
  Split2(tlm, tl, tm, x);
  int ans = tm->val;
  t = Merge(tl, Merge(tm, tr));
  return ans;
}
int main() {
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    b[i] = {a, i};
  }
  sort(b + 1, b + n + 1, cmp);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int k, pos;
    cin >> k >> pos;
    q[i] = {{k, pos}, i};
  }
  sort(q + 1, q + m + 1, cmp2);
  for (int i = 1; i <= m; i++) {
    int all = q[i].first.first, pos = q[i].first.second, id = q[i].second,
        prev = q[i - 1].first.first;
    if (all > prev) {
      for (int j = prev + 1; j <= all; j++) {
        Insert(t, b[j]);
      }
    }
    ans[id] = Find(t, pos);
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
struct node {
  int val, pos, sz, pr;
  node *l, *r;
};
node *t;
int ans[200020];
pair<int, int> b[200020];
pair<pair<int, int>, int> q[200020];
bool cmp(pair<int, int> x, pair<int, int> y) {
  return (x.first > y.first) || (x.first == y.first && x.second < y.second);
}
bool cmp2(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  return (x.first.first < y.first.first) ||
         (x.first.first == y.first.first && x.first.second < y.first.second);
}
int get_size(node *t) { return (t == 0) ? 0 : t->sz; }
void Upd(node *t) {
  if (t != 0) t->sz = get_size(t->l) + get_size(t->r) + 1;
}
node *Merge(node *tl, node *tr) {
  if (tl == 0) return tr;
  if (tr == 0) return tl;
  if (tl->pr < tr->pr) {
    tr->l = Merge(tl, tr->l);
    Upd(tr);
    return tr;
  } else {
    tl->r = Merge(tl->r, tr);
    Upd(tl);
    return tl;
  }
}
void Split(node *t, node *&tl, node *&tr, int x) {
  if (t == 0) {
    tl = tr = 0;
    return;
  }
  if (t->pos < x) {
    Split(t->r, t->r, tr, x);
    tl = t;
    Upd(t);
  } else {
    Split(t->l, tl, t->l, x);
    tr = t;
    Upd(t);
  }
}
void Split2(node *t, node *&tl, node *&tr, int x) {
  if (t == 0) {
    tl = tr = 0;
    return;
  }
  if (get_size(t->l) + 1 < x) {
    int newx = x - get_size(t->l) - 1;
    Split2(t->r, t->r, tr, newx);
    tl = t;
    Upd(t);
  } else {
    Split2(t->l, tl, t->l, x);
    tr = t;
    Upd(t);
  }
}
void Out(node *t) {
  if (t == 0) return;
  Out(t->l);
  cout << t->val << " ";
  Out(t->r);
}
void Insert(node *&t, pair<int, int> x) {
  node *tl, *tmr, *tm, *tr;
  Split(t, tl, tmr, x.second);
  Split(tmr, tm, tr, x.second + 1);
  tm = new node;
  tm->pos = x.second;
  tm->pr = rand() * rand();
  tm->sz = 1;
  tm->val = x.first;
  tm->l = tm->r = 0;
  t = Merge(tl, Merge(tm, tr));
}
int Find(node *&t, int x) {
  node *tlm, *tr, *tl, *tm;
  Split2(t, tlm, tr, x + 1);
  Split2(tlm, tl, tm, x);
  int ans = tm->val;
  t = Merge(tl, Merge(tm, tr));
  return ans;
}
int main() {
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    b[i] = {a, i};
  }
  sort(b + 1, b + n + 1, cmp);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int k, pos;
    cin >> k >> pos;
    q[i] = {{k, pos}, i};
  }
  sort(q + 1, q + m + 1, cmp2);
  for (int i = 1; i <= m; i++) {
    int all = q[i].first.first, pos = q[i].first.second, id = q[i].second,
        prev = q[i - 1].first.first;
    if (all > prev) {
      for (int j = prev + 1; j <= all; j++) {
        Insert(t, b[j]);
      }
    }
    ans[id] = Find(t, pos);
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
  return 0;
}

```