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: // Don't place your source in a package
  
  
  import javax.swing.*;
  import java.lang.reflect.Array;
  import java.text.DecimalFormat;
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  import java.math.*;
  import java.util.stream.Stream;
  
  
  
  // Please name your class Main
  public class Main {
      static FastScanner fs=new FastScanner();
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          public String next() {
              while (!st.hasMoreElements())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
          int Int() {
              return Integer.parseInt(next());
          }
  
          long Long() {
              return Long.parseLong(next());
          }
  
          String Str(){
              return next();
          }
      }
  
  
      public static void main (String[] args) throws java.lang.Exception {
          PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
          //reading /writing file
          //Scanner sc=new Scanner(new File("input.txt"));
          //PrintWriter pr=new PrintWriter("output.txt");
  
  
          int T=Int();
          for(int t=0;t<T;t++){
              int n=Int();
              int k=Int();
              int A[]=Arr(n);
              Solution sol=new Solution(out);
              sol.solution(A,k);
          }
          out.close();
  
      }
  
  
  
  
      public static int[] Arr(int n){
          int A[]=new int[n];
          for(int i=0;i<n;i++)A[i]=Int();
          return A;
      }
      public static int Int(){
          return fs.Int();
      }
      public static long Long(){
          return fs.Long();
      }
      public static String Str(){
          return fs.Str();
      }
  
  }
  
  
  
  
  
  
  
  
  class Solution {
      PrintWriter out;
      int INF = Integer.MAX_VALUE;
      int MOD = 1000000007;
      int mod = MOD;
      public Solution(PrintWriter out) {
          this.out = out;
      }
  
  
  
  
      public void solution(int A[],int k) {
          int dif[]=new int[A.length];
          for(int i=0;i<A.length;i++){
              dif[i]=A[i]-(1+i);
          }
  
          int l =0,r=A.length-1;
          int res=-1;
          while(l<=r){
              int mid = l+(r-l)/2;
              int cnt =get(dif,mid);
              if(cnt>=k){
                  res=mid;
                  r=mid-1;
              }
              else{
                  l=mid+1;
              }
          }
          out.println(res);
      }
  
      public int get(int A[],int total){
          int dp[][]=new int[A.length][total+1];
          for(int i=0;i<dp.length;i++){
              Arrays.fill(dp[i],-1);
          }
          int res= dfs(dp,A,0,0,total);
          return res;
      }
      public int dfs(int dp[][],int A[],int i,int cnt,int total){
          if(i>=A.length){
              return 0;
          }
          if(dp[i][cnt]!=-1)return dp[i][cnt];
          int res = 0;
          if(cnt<total){
              int a = 0;
              int b=0;
  
              if(A[i]+cnt==0){
                  a++;
              }
              a+=dfs(dp,A,i+1,cnt,total);
  
  
              b+=dfs(dp,A,i+1,cnt+1,total);
              res=Math.max(a,b);
  
          }
          else{
              if(A[i]+cnt==0)res++;
              res+=dfs(dp,A,i+1,cnt,total);
          }
          dp[i][cnt]=res;
          return res;
      }
  
  
  
  
  }
  