description: Consider a sequence of integers a_1, a_2, …, a_n. In one move, you can select any element of the sequence and delete it. After an element is deleted, all elements to the right are shifted to the left by 1 position, so there are no empty spaces in the sequence. So after you make a move, the sequence's length decreases by 1. The indices of the elements after the move are recalculated.
  
  E. g. let the sequence be a=[3, 2, 2, 1, 5]. Let's select the element a_3=2 in a move. Then after the move the sequence will be equal to a=[3, 2, 1, 5], so the 3-rd element of the new sequence will be a_3=1 and the 4-th element will be a_4=5.
  
  You are given a sequence a_1, a_2, …, a_n and a number k. You need to find the minimum number of moves you have to make so that in the resulting sequence there will be at least k elements that are equal to their indices, i. e. the resulting sequence b_1, b_2, …, b_m will contain at least k indices i such that b_i = i.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
  
  Each test case consists of two consecutive lines. The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2000). The second line contains a sequence of integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). The numbers in the sequence are not necessarily different.
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 2000.
  
  Output
  
  For each test case output in a single line:
  
    * -1 if there's no desired move sequence; 
    * otherwise, the integer x (0 ≤ x ≤ n) — the minimum number of the moves to be made so that the resulting sequence will contain at least k elements that are equal to their indices. 
  
  Example
  
  Input
  
  
  4
  7 6
  1 1 2 3 4 5 6
  5 2
  5 1 3 2 3
  5 2
  5 5 5 5 4
  8 4
  1 2 3 3 2 2 5 5
  
  
  Output
  
  
  1
  2
  -1
  2
  
  Note
  
  In the first test case the sequence doesn't satisfy the desired condition, but it can be provided by deleting the first element, hence the sequence will be [1, 2, 3, 4, 5, 6] and 6 elements will be equal to their indices.
  
  In the second test case there are two ways to get the desired result in 2 moves: the first one is to delete the 1-st and the 3-rd elements so that the sequence will be [1, 2, 3] and have 3 elements equal to their indices; the second way is to delete the 2-nd and the 3-rd elements to get the sequence [5, 2, 3] with 2 desired elements.
solution: #include <bits/stdc++.h>
  using namespace std;
  template <typename T>
  ostream& operator<<(ostream& os, vector<T>& v) {
    if (v.size() == 0) {
      os << "empty vector\n";
      return os;
    }
    for (auto element : v) os << element << " ";
    return os;
  }
  template <typename T, typename second>
  ostream& operator<<(ostream& os, pair<T, second>& p) {
    os << "(" << p.first << ", " << p.second << ")";
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, set<T>& v) {
    if (v.size() == 0) {
      os << "empty set\n";
      return os;
    }
    auto endit = v.end();
    endit--;
    os << "[";
    for (auto it = v.begin(); it != v.end(); it++) {
      os << *it;
      if (it != endit) os << ", ";
    }
    os << "]";
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, multiset<T>& v) {
    if (v.size() == 0) {
      os << "empty multiset\n";
      return os;
    }
    auto endit = v.end();
    endit--;
    os << "[";
    for (auto it = v.begin(); it != v.end(); it++) {
      os << *it;
      if (it != endit) os << ", ";
    }
    os << "]";
    return os;
  }
  template <typename T, typename second>
  ostream& operator<<(ostream& os, map<T, second>& v) {
    if (v.size() == 0) {
      os << "empty map\n";
      return os;
    }
    auto endit = v.end();
    endit--;
    os << "{";
    for (auto it = v.begin(); it != v.end(); it++) {
      os << "(" << (*it).first << " : " << (*it).second << ")";
      if (it != endit) os << ", ";
    }
    os << "}";
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, vector<vector<T>>& v) {
    for (auto& subv : v) {
      for (auto& e : subv) os << e << " ";
      os << "\n";
    }
    return os;
  }
  bool do_debug = false;
  long long n, k;
  vector<long long> v;
  long long dp[2010][2010];
  long long solve(long long index, long long done, long long maxi) {
    if (index == n) {
      if (v[index] == index - done)
        return 1;
      else
        return 0;
    }
    if (dp[index][done] != -1) return dp[index][done];
    if (done == maxi) {
      if (v[index] == index - done)
        return dp[index][done] = solve(index + 1, done, maxi) + 1;
      else
        return dp[index][done] = solve(index + 1, done, maxi);
    }
    long long tem1 = solve(index + 1, done + 1, maxi);
    long long tem2;
    if (v[index] == index - done) {
      tem2 = solve(index + 1, done, maxi) + 1;
    } else
      tem2 = solve(index + 1, done, maxi);
    return dp[index][done] = max(tem1, tem2);
  }
  long long max_val = 0;
  bool f(long long mid) {
    for (long long i = 0; i < n + 2; i++)
      for (long long j = 0; j < mid + 2; j++) dp[i][j] = -1;
    long long tem = solve(1, 0, mid);
    if (tem < k)
      return false;
    else
      return true;
  }
  void Runtime_Terror() {
    cin >> n >> k;
    v.resize(n + 1);
    for (long long i = 1; i < n + 1; i++) cin >> v[i];
    long long l = 0, r = n;
    if (f(0)) {
      cout << 0 << "\n";
      return;
    }
    if (f(n) == 0) {
      cout << -1 << "\n";
      return;
    }
    while (l <= r) {
      long long mid = (l + r) / 2;
      bool b1 = f(mid), b2 = f(mid + 1);
      if (!b1 && b2) {
        cout << mid + 1 << "\n";
        return;
      }
      if (b1)
        r = mid;
      else
        l = mid + 1;
    }
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    for (long long i = 0; i < t; i++) Runtime_Terror();
    return 0;
  }
  