#include <bits/stdc++.h>
using namespace std;
using VP = vector<pair<long long, long long> >;
using VVP = vector<VP>;
using VI = vector<long long>;
using VVI = vector<VI>;
using VVVI = vector<VVI>;
const int inf = 1e9 + 7;
const long long INF = 1LL << 58;
const long long mod = 1e9 + 7;
template <class Abel>
struct BIT {
  vector<Abel> dat;
  Abel UNITY_SUM = 0;
  BIT(int n) { init(n); }
  void init(int n) {
    dat.resize(n + 1);
    for (int i = 0; i < (int)dat.size(); ++i) dat[i] = UNITY_SUM;
  }
  inline void add(int a, Abel x) {
    for (int i = a; i < (int)dat.size(); i += i & -i) dat[i] = dat[i] + x;
  }
  inline Abel sum(int a) {
    Abel res = UNITY_SUM;
    for (int i = a; i > 0; i -= i & -i) res = res + dat[i];
    return res;
  }
  inline Abel sum(int a, int b) { return sum(b - 1) - sum(a - 1); }
  int get(long long k) {
    ++k;
    int res = 0;
    int N = 1;
    while (N < (int)dat.size()) N *= 2;
    for (int i = N / 2; i > 0; i /= 2) {
      if (res + i < (int)dat.size() && dat[res + i] < k) {
        k = k - dat[res + i];
        res = res + i;
      }
    }
    return res + 1;
  }
};
int main() {
  int i, j;
  long long n;
  cin >> n;
  VI a(n);
  VI b;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    a[i] *= -1;
    b.push_back(a[i]);
  }
  sort(b.begin(), b.end());
  b.erase(unique(b.begin(), b.end()), b.end());
  int m;
  cin >> m;
  map<long long, long long> ma;
  for (i = 0; i < n; i++) {
    ma[a[i]]++;
  }
  map<long long, long long> m2 = ma;
  long long sum = 0;
  for (auto it = m2.begin(); it != m2.end(); ++it) {
    sum += it->second;
    it->second = sum;
  }
  long long c[n + 1];
  long long d[n + 1];
  map<long long, long long> cnt;
  for (j = 0; j < n; j++) {
    cnt[a[j]]++;
    c[j] = m2[a[j]] - ma[a[j]] + cnt[a[j]];
    d[c[j]] = j;
  }
  for (i = 0; i <= n; i++) {
  }
  vector<pair<pair<long long, long long>, long long> > an;
  for (i = 0; i < m; i++) {
    long long k, p;
    cin >> k >> p;
    an.push_back(make_pair(pair<long long, long long>(k, i), p));
  }
  sort(an.begin(), an.end());
  long long ans[m + 1];
  BIT<int> bit(200020);
  long long r = 1;
  for (int i = 0; i < m; i++) {
    while (r <= an[i].first.first) {
      bit.add(d[r] + 1, 1);
      r++;
    }
    long long val = bit.get(an[i].second - 1) - 1;
    ans[an[i].first.second] = -a[val];
  }
  for (i = 0; i < m; i++) {
    cout << ans[i] << endl;
  }
}

### 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;
using VP = vector<pair<long long, long long> >;
using VVP = vector<VP>;
using VI = vector<long long>;
using VVI = vector<VI>;
using VVVI = vector<VVI>;
const int inf = 1e9 + 7;
const long long INF = 1LL << 58;
const long long mod = 1e9 + 7;
template <class Abel>
struct BIT {
  vector<Abel> dat;
  Abel UNITY_SUM = 0;
  BIT(int n) { init(n); }
  void init(int n) {
    dat.resize(n + 1);
    for (int i = 0; i < (int)dat.size(); ++i) dat[i] = UNITY_SUM;
  }
  inline void add(int a, Abel x) {
    for (int i = a; i < (int)dat.size(); i += i & -i) dat[i] = dat[i] + x;
  }
  inline Abel sum(int a) {
    Abel res = UNITY_SUM;
    for (int i = a; i > 0; i -= i & -i) res = res + dat[i];
    return res;
  }
  inline Abel sum(int a, int b) { return sum(b - 1) - sum(a - 1); }
  int get(long long k) {
    ++k;
    int res = 0;
    int N = 1;
    while (N < (int)dat.size()) N *= 2;
    for (int i = N / 2; i > 0; i /= 2) {
      if (res + i < (int)dat.size() && dat[res + i] < k) {
        k = k - dat[res + i];
        res = res + i;
      }
    }
    return res + 1;
  }
};
int main() {
  int i, j;
  long long n;
  cin >> n;
  VI a(n);
  VI b;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    a[i] *= -1;
    b.push_back(a[i]);
  }
  sort(b.begin(), b.end());
  b.erase(unique(b.begin(), b.end()), b.end());
  int m;
  cin >> m;
  map<long long, long long> ma;
  for (i = 0; i < n; i++) {
    ma[a[i]]++;
  }
  map<long long, long long> m2 = ma;
  long long sum = 0;
  for (auto it = m2.begin(); it != m2.end(); ++it) {
    sum += it->second;
    it->second = sum;
  }
  long long c[n + 1];
  long long d[n + 1];
  map<long long, long long> cnt;
  for (j = 0; j < n; j++) {
    cnt[a[j]]++;
    c[j] = m2[a[j]] - ma[a[j]] + cnt[a[j]];
    d[c[j]] = j;
  }
  for (i = 0; i <= n; i++) {
  }
  vector<pair<pair<long long, long long>, long long> > an;
  for (i = 0; i < m; i++) {
    long long k, p;
    cin >> k >> p;
    an.push_back(make_pair(pair<long long, long long>(k, i), p));
  }
  sort(an.begin(), an.end());
  long long ans[m + 1];
  BIT<int> bit(200020);
  long long r = 1;
  for (int i = 0; i < m; i++) {
    while (r <= an[i].first.first) {
      bit.add(d[r] + 1, 1);
      r++;
    }
    long long val = bit.get(an[i].second - 1) - 1;
    ans[an[i].first.second] = -a[val];
  }
  for (i = 0; i < m; i++) {
    cout << ans[i] << endl;
  }
}

```