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.lang.*;
  import java.util.*;
  import java.io.*;
  import java.math.*;
  
  public class Main {
      static void deal(int n,int k,int[] arr) {
          int[] a = new int[n];
          for(int i=0;i<n;i++) {
              a[i] = i+1-arr[i];
          }
  //        out.println(Arrays.toString(a));
          int[][] dp = new int[n+1][n+1];
          dp[0][0] = a[0]==0?1:0;
          for(int i=1;i<n;i++) {
              dp[0][i] = dp[0][i-1]+(a[i]==0?1:0);
          }
          int j=0;
  //        out.println(Arrays.toString(dp[0]));
          while(dp[j][n-1]<k) {
              j++;
              if(j==n) break;
              dp[j][j] = a[j]==j?1:0;
              for(int i=j+1;i<n;i++) {
                  dp[j][i] = Math.max(dp[j-1][i-1]-(a[i-1]==j-1?1:0),dp[j][i-1])+(a[i]==j?1:0);
              }
  //            out.println(Arrays.toString(dp[j]));
          }
          if(j==n) {
              out.println(-1);
          } else {
              out.println(j);
          }
      }
      
  	public static void main(String[] args) {
          MyScanner sc = new MyScanner();
          out = new PrintWriter(new BufferedOutputStream(System.out));
          int t = sc.nextInt();
          for(int i=0;i<t;i++) {
              int n = sc.nextInt();
              int k = sc.nextInt();
              int[] arr = new int[n];
              for(int j=0;j<n;j++) arr[j] = sc.nextInt();
              deal(n,k,arr);
          }
          out.close();
      }
      
      //-----------PrintWriter for faster output---------------------------------
      public static PrintWriter out;
      
      //-----------MyScanner class for faster input----------
      public static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
          
          public MyScanner() {
                  br = new BufferedReader(new InputStreamReader(System.in));
  //            try {
  //                br = new BufferedReader(new InputStreamReader(new FileInputStream("/Users/frank/Downloads/test.txt")));
  //            } catch (Exception e) {
  //                
  //            }
  
          }
          
          String next() {
                  while (st == null || !st.hasMoreElements()) {
                          try {
                                  st = new StringTokenizer(br.readLine());
                          } catch (IOException e) {
                                  e.printStackTrace();
                          }
                  }
                  return st.nextToken();
          }
          
          int nextInt() {
                  return Integer.parseInt(next());
          }
          
          long nextLong() {
                  return Long.parseLong(next());
          }
          
          double nextDouble() {
                  return Double.parseDouble(next());
          }
          
          String nextLine(){
                  String str = "";
      try {
              str = br.readLine();
      } catch (IOException e) {
              e.printStackTrace();
      }
      return str;
          }
          
      }
  }