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: import java.io.*;
  import java.util.*;
  import java.lang.*;
  
  public class E {
    public static void main(String[] args) {
      InputStream inputStream = System.in;
      OutputStream outputStream = System.out;
      InputReader in = new InputReader(inputStream);
      PrintWriter out = new PrintWriter(outputStream);
      Solver solver = new Solver();
      solver.Main(in, out);
      out.close();
    }
  
    static class Solver {
      public void Main(InputReader in, PrintWriter out) {
        int T = in.nextInt();
        for (int t = 0; t < T; t++) {
          int n = in.nextInt();
          int k = in.nextInt();
          int[] a = new int[n];
          for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
            a[i] -= 1;
          }
          int[][] dp = new int[n + 1][n + 1];
          for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= n; j++) {
              dp[i][j] = (int) -1e9;
            }
          }
          dp[0][0] = 0;
          for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
              int add = 0;
              if (a[i] == j) add = 1;
              dp[i + 1][j] = Math.max(dp[i + 1][j], dp[i][j]);
              dp[i + 1][j + 1] = Math.max(dp[i + 1][j + 1], dp[i][j] + add);
            }
          }
          int ans = -1;
          for (int i = 1; i <= n; i++) {
            if (dp[n][i] >= k) {
              ans = Math.max(ans, i);
            }
          }
          if (ans != -1) {
            ans = n - ans;
          }
          out.println(ans);
        }
      }
    }
  
    static class InputReader {
      public BufferedReader reader;
      public StringTokenizer tokenizer;
  
      public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
      }
  
      public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
          try {
            tokenizer = new StringTokenizer(reader.readLine());
          } catch (IOException e) {
            throw new RuntimeException(e);
          }
        }
        return tokenizer.nextToken();
      }
  
      public int nextInt() {
        return Integer.parseInt(next());
      }
    }
  }
  