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.util.*;
  import java.io.*;
  
  public class Main {
      public static void main(String args[]) {new Main().run();}
  
      FastReader in = new FastReader();
      PrintWriter out = new PrintWriter(System.out);
      void run() {
          for(int q=ni();q>0;q--){
              work();
          }
          out.flush();
      }
      long mod=1000000007;
      long gcd(long a,long b) {
          return a==0?b:gcd(b%a,a);
      }
      void work() {
          int n=ni(),k=ni();
          int[][] dp=new int[n+1][n+1];
          int[] A=new int[n+1];
          for(int i=1;i<=n;i++){
              A[i]=ni();
          }
          for(int i=1;i<=n;i++){
              for(int j=0;j<=i;j++){
                  dp[i][j]=dp[i-1][j]+(A[i]==i-j?1:0);
                  if(j>0){
                      dp[i][j]=Math.max(dp[i][j],dp[i-1][j-1]);
                  }
              }
          }
          for(int j=0;j<=n;j++){
              if(dp[n][j]>=k){
                  out.println(j);
                  return;
              }
          }
          out.println(-1);
      }
  
      //input
      @SuppressWarnings("unused")
      private ArrayList<Integer>[] ng(int n, int m) {
          ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
          for(int i=0;i<n;i++) {
              graph[i]=new ArrayList<>();
          }
          for(int i=1;i<=m;i++) {
              int s=in.nextInt()-1,e=in.nextInt()-1;
              graph[s].add(e);
              graph[e].add(s);
          }
          return graph;
      }
  
      private ArrayList<long[]>[] ngw(int n, int m) {
          ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
          for(int i=0;i<n;i++) {
              graph[i]=new ArrayList<>();
          }
          for(int i=1;i<=m;i++) {
              long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
              graph[(int)s].add(new long[] {e,w});
              graph[(int)e].add(new long[] {s,w});
          }
          return graph;
      }
  
      private int ni() {
          return in.nextInt();
      }
  
      private long nl() {
          return in.nextLong();
      }
      private double nd() {
          return in.nextDouble();
      }
      private String ns() {
          return in.next();
      }
  
      private long[] na(int n) {
          long[] A=new long[n];
          for(int i=0;i<n;i++) {
              A[i]=in.nextLong();
          }
          return A;
      }
  
      private int[] nia(int n) {
          int[] A=new int[n];
          for(int i=0;i<n;i++) {
              A[i]=in.nextInt();
          }
          return A;
      }
  }
  
  class FastReader
  {
      BufferedReader br;
      StringTokenizer st;
      InputStreamReader input;//no buffer
      public FastReader()
      {
          br=new BufferedReader(new InputStreamReader(System.in));
      }
  
      public FastReader(boolean isBuffer)
      {
          if(!isBuffer){
              input=new InputStreamReader(System.in);
          }else{
              br=new BufferedReader(new InputStreamReader(System.in));
          }
      }
  
      public boolean hasNext(){
          try{
              String s=br.readLine();
              if(s==null){
                  return  false;
              }
              st=new StringTokenizer(s);
          }catch(IOException e){
              e.printStackTrace();
          }
          return true;
      }
  
      public String next()
      {   
          if(input!=null){
              try {
                  StringBuilder sb=new StringBuilder();
                  int ch=input.read();
                  while(ch=='\n'||ch=='\r'||ch==32){
                      ch=input.read();
                  }
                  while(ch!='\n'&&ch!='\r'&&ch!=32){
                      sb.append((char)ch);
                      ch=input.read();
                  }
                  return sb.toString();
              }catch (Exception e){
                  e.printStackTrace();
              }
          }
          while(st==null || !st.hasMoreElements())//回车，空行情况
          {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
  
      public int nextInt()
      {
          return (int)nextLong();
      }
  
      public long nextLong() {
          try {
              if(input!=null){
                  long ret=0;
                  int b=input.read();
                  while(b<'0'||b>'9'){
                      b=input.read();
                  }
                  while(b>='0'&&b<='9'){
                      ret=ret*10+(b-'0');
                      b=input.read();
                  }
                  return ret;
              }
          }catch (Exception e){
              e.printStackTrace();
          }
          return Long.parseLong(next());
      }
  
      public double nextDouble()
      {
          return Double.parseDouble(next());
      }
  }