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;
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  double EPS = 1e-9;
  int INF = 1000000005;
  long long INFF = 1000000000000000005ll;
  double PI = acos(-1);
  int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
  int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
  template <typename T>
  int SIZE(T(&t)) {
    return t.size();
  }
  template <typename T, size_t N>
  int SIZE(T (&t)[N]) {
    return N;
  }
  string to_string(char t) { return "'" + string({t}) + "'"; }
  string to_string(bool t) { return t ? "true" : "false"; }
  string to_string(const string &t, int x1 = 0, int x2 = 1e9) {
    string ret = "";
    for (int i = min(x1, SIZE(t)), _i = min(x2, SIZE(t) - 1); i <= _i; ++i) {
      ret += t[i];
    }
    return '"' + ret + '"';
  }
  string to_string(const char *t) {
    string ret(t);
    return to_string(ret);
  }
  template <size_t N>
  string to_string(const bitset<N> &t, int x1 = 0, int x2 = 1e9) {
    string ret = "";
    for (int i = min(x1, SIZE(t)); i <= min(x2, SIZE(t) - 1); ++i) {
      ret += t[i] + '0';
    }
    return to_string(ret);
  }
  template <typename T, typename... Coords>
  string to_string(const T(&t), int x1 = 0, int x2 = 1e9, Coords... C);
  template <typename T, typename S>
  string to_string(const pair<T, S> &t) {
    return "(" + to_string(t.first) + ", " + to_string(t.second) + ")";
  }
  template <typename T, typename... Coords>
  string to_string(const T(&t), int x1, int x2, Coords... C) {
    string ret = "[";
    x1 = min(x1, SIZE(t));
    auto e = begin(t);
    advance(e, x1);
    for (int i = x1, _i = min(x2, SIZE(t) - 1); i <= _i; ++i) {
      ret += to_string(*e, C...) + (i != _i ? ", " : "");
      e = next(e);
    }
    return ret + "]";
  }
  template <int Index, typename... Ts>
  struct print_tuple {
    string operator()(const tuple<Ts...> &t) {
      string ret = print_tuple<Index - 1, Ts...>{}(t);
      ret += (Index ? ", " : "");
      return ret + to_string(get<Index>(t));
    }
  };
  template <typename... Ts>
  struct print_tuple<0, Ts...> {
    string operator()(const tuple<Ts...> &t) { return to_string(get<0>(t)); }
  };
  template <typename... Ts>
  string to_string(const tuple<Ts...> &t) {
    const auto Size = tuple_size<tuple<Ts...>>::value;
    return print_tuple<Size - 1, Ts...>{}(t);
  }
  void dbgr() { ; }
  template <typename Heads, typename... Tails>
  void dbgr(Heads H, Tails... T) {
    cout << to_string(H) << " | ";
    dbgr(T...);
  }
  void dbgs() { ; }
  template <typename Heads, typename... Tails>
  void dbgs(Heads H, Tails... T) {
    cout << H << " ";
    dbgs(T...);
  }
  const int MOD = 1000000007;
  template <typename T_vector>
  void output_vector(const T_vector &v, bool line_break = false,
                     bool add_one = false, int start = -1, int end = -1) {
    if (start < 0) start = 0;
    if (end < 0) end = int(v.size());
    for (int i = start; i < end; i++) {
      cout << v[i] + (add_one ? 1 : 0)
           << (line_break    ? '\n'
               : i < end - 1 ? ' '
                             : '\n');
    }
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n);
    for (int(i) = (0); (i) < (n); ++(i)) {
      cin >> a[i];
      --a[i];
    }
    vector<vector<long long>> dp(n + 1, vector<long long>(n, -INFF));
    dp[0][0] = 0;
    for (int(i) = (0); (i) < (n); ++(i)) {
      for (int(j) = (0); (j) < (n); ++(j)) {
        if (dp[i][j] < 0) continue;
        dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + (a[i] == i - j));
        if (j + 1 < n) dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
      }
    }
    long long ans = n;
    for (int(j) = (0); (j) < (n); ++(j)) {
      if (dp[n][j] >= k) {
        ans = min(ans, (long long)j);
      }
    }
    cout << (ans == n ? -1 : ans) << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int tc;
    cin >> tc;
    while (tc--) solve();
    return 0;
  }
  