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.lang.*;
  import java.io.*;
  
  //           THIS TEMPLATE MADE BY AKSH BANSAL.
  
  public class Solution {
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          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;
          }
      }
      private static boolean[] isPrime;
      private static void primes(){
          int num = (int)1e6; // PRIMES FROM 1 TO NUM
          isPrime = new boolean[num];
       
          for (int i = 2; i< isPrime.length; i++) {
             isPrime[i] = true;
          }
          for (int i = 2; i< Math.sqrt(num); i++) {
             if(isPrime[i] == true) {
                for(int j = (i*i); j<num; j = j+i) {
                   isPrime[j] = false;
                }
             }
          }
      }
      static void sort(int a[]){ // int -> long
          ArrayList<Integer> arr=new ArrayList<>(); // Integer -> Long
          for(int i=0;i<a.length;i++)
          arr.add(a[i]);
          Collections.sort(arr);
          for(int i=0;i<a.length;i++)
          a[i]=arr.get(i);
          
      }
      private static long gcd(long a, long b){
          if(b==0)return a;
          return gcd(b,a%b);
      }
      private static long pow(long x,long y){
          if(y==0)return 1;
          long temp = pow(x, y/2);
          if(y%2==1){
              return x*temp*temp;
          }
          else{
              return temp*temp;
          }
      }
      // static ArrayList<Integer>[] adj;
      // static void getAdj(int n,int q, FastReader sc){
      //     adj = new ArrayList[n+1];
      //     for(int i=1;i<=n;i++){
      //         adj[i] = new ArrayList<>();
      //     }
      //     for(int i=0;i<q;i++){
      //         int a = sc.nextInt();
      //         int b = sc.nextInt();
  
      //         adj[a].add(b);
      //         adj[b].add(a);
      //     }
      // }
      static PrintWriter out;
      static FastReader sc ;
      public static void main(String[] args) throws IOException {
          sc = new FastReader();
          out = new PrintWriter(System.out);
          // primes();
          // ________________________________
  
          int t = sc.nextInt();
          StringBuilder output = new StringBuilder();
  
          while (t-- > 0) {
              int n = sc.nextInt();
              int k = sc.nextInt();
              int[] arr = new int[n];
              for(int i=0;i<n;i++){
                  arr[i] = sc.nextInt();
              }
              output.append(solver(arr, n, k)).append("\n");
          }
  
          out.println(output);
          // _______________________________
  
          // int n = sc.nextInt();
          // out.println(solver());
          // ________________________________
          out.flush();
      }
  
      public static int solver(int[] arr, int n, int k) {
          int[][] dp = new int[n+1][n+1];
  
          for(int i=1;i<=n;i++){
              for(int j=0;j<=i;j++){
                  int cur = 0;
                  if(arr[i-1]+j==i){
                      cur++;
                  }
                  if(j==0){
                      dp[i][j] = dp[i-1][j]+cur;
                  }
                  else {
                      dp[i][j] = Math.max(dp[i-1][j]+cur, dp[i-1][j-1]);
                  }
              }
          }
          for(int i=0;i<=n;i++){
              if(dp[n][i]>=k){
                  return i;
              }
          }
          return -1;
      }
  }