#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 7;
int n, m, w[N], val[N], vat[N];
struct node {
  int a, k;
} f[N];
struct query {
  int t, k, c, s;
} e[N];
inline void add(int u, int l, int r, int t, int xt) {
  if (!vat[t]) vat[t] = xt;
  vat[t] = min(vat[t], xt);
  val[t]++;
  if (l == r) return;
  int d = (l + r) >> 1;
  if (u <= d)
    add(u, l, d, t << 1, xt);
  else
    add(u, d + 1, r, t << 1 | 1, xt);
}
inline int getans(int u, int l, int r, int t, int q) {
  if (l == r) return l;
  int d = (l + r) >> 1;
  if (q + val[t << 1] >= u) return getans(u, l, d, t << 1, q);
  return getans(u, d + 1, r, t << 1 | 1, q + val[t << 1]);
}
inline int getit(int ql, int qr, int l, int r, int t) {
  if (ql == l && qr == r) return vat[t];
  int d = (l + r) >> 1, vas = 1e9 + 7;
  if (ql <= d) vas = min(vas, getit(ql, min(qr, d), l, d, t << 1));
  if (d + 1 <= qr)
    vas = min(vas, getit(max(d + 1, ql), qr, d + 1, r, t << 1 | 1));
  return vas;
}
inline bool cmp(node a, node b) {
  if (a.a == b.a) return a.k < b.k;
  return a.a > b.a;
}
inline bool tmp(query a, query b) { return a.s < b.s; }
inline bool qmp(query a, query b) { return a.c < b.c; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &f[i].a), f[i].k = i, w[i] = f[i].a;
  sort(f + 1, f + n + 1, cmp);
  cin >> m;
  for (int i = 1; i <= m; i++) scanf("%d%d", &e[i].s, &e[i].t), e[i].c = i;
  sort(e + 1, e + m + 1, tmp);
  int c = 1;
  for (int i = 1; i <= m; i++) {
    while (c <= e[i].s) add(f[c].k, 1, n, 1, f[c].a), c++;
    e[i].t = getans(e[i].t, 1, n, 1, 0);
  }
  sort(e + 1, e + m + 1, qmp);
  for (int i = 1; i <= m; i++) printf("%d\n", w[e[i].t]);
}

### 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 int N = 1e6 + 7;
int n, m, w[N], val[N], vat[N];
struct node {
  int a, k;
} f[N];
struct query {
  int t, k, c, s;
} e[N];
inline void add(int u, int l, int r, int t, int xt) {
  if (!vat[t]) vat[t] = xt;
  vat[t] = min(vat[t], xt);
  val[t]++;
  if (l == r) return;
  int d = (l + r) >> 1;
  if (u <= d)
    add(u, l, d, t << 1, xt);
  else
    add(u, d + 1, r, t << 1 | 1, xt);
}
inline int getans(int u, int l, int r, int t, int q) {
  if (l == r) return l;
  int d = (l + r) >> 1;
  if (q + val[t << 1] >= u) return getans(u, l, d, t << 1, q);
  return getans(u, d + 1, r, t << 1 | 1, q + val[t << 1]);
}
inline int getit(int ql, int qr, int l, int r, int t) {
  if (ql == l && qr == r) return vat[t];
  int d = (l + r) >> 1, vas = 1e9 + 7;
  if (ql <= d) vas = min(vas, getit(ql, min(qr, d), l, d, t << 1));
  if (d + 1 <= qr)
    vas = min(vas, getit(max(d + 1, ql), qr, d + 1, r, t << 1 | 1));
  return vas;
}
inline bool cmp(node a, node b) {
  if (a.a == b.a) return a.k < b.k;
  return a.a > b.a;
}
inline bool tmp(query a, query b) { return a.s < b.s; }
inline bool qmp(query a, query b) { return a.c < b.c; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &f[i].a), f[i].k = i, w[i] = f[i].a;
  sort(f + 1, f + n + 1, cmp);
  cin >> m;
  for (int i = 1; i <= m; i++) scanf("%d%d", &e[i].s, &e[i].t), e[i].c = i;
  sort(e + 1, e + m + 1, tmp);
  int c = 1;
  for (int i = 1; i <= m; i++) {
    while (c <= e[i].s) add(f[c].k, 1, n, 1, f[c].a), c++;
    e[i].t = getans(e[i].t, 1, n, 1, 0);
  }
  sort(e + 1, e + m + 1, qmp);
  for (int i = 1; i <= m; i++) printf("%d\n", w[e[i].t]);
}

```