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 Sol{
  
  /*   
  ->n=1 important test case
  ->Dont get stuck on one approach try all possibilities
  ->check int overflow,array bounds,etc
  ->Dont think just idea think of its (correct) implementation completely (then only code)
  */
  
  public static void main(String []args){
  //precomp();
  int times=ni();while(times-->0){solve();}out.close();}
  
  static void solve(){  
     int n=ni();int k=ni();
     int a[]=new int[n+1];
     int del[]=new int[n+1];
     for(int i=1;i<=n;i++){a[i]=ni();del[i]=i-a[i];}//out.print(del[i]+" ");}out.println();
  
     int dp[]=new int[n+1];
     for(int i=1;i<=n;i++){
        if(del[i]>=0){
          for(int j=0;j<i;j++){
              if(del[j]<0){}
              else{
                if(del[i]>=del[j] && del[i]<=del[j]+i-j-1){dp[i]=Math.max(dp[i],dp[j]);}
              }
           }
           dp[i]++;
        }
        else{dp[i]=-1;}
        //out.print(dp[i]+" ");
     
     }
     //out.println();
     for(int i=1;i<=n;i++){
       for(int j=i+1;j<=n;j++){
           if(del[j]==del[i])dp[i]++;
       }//out.print(dp[i]+" ");
     }
     //out.println();
     int ans=Max;
     for(int i=0;i<=n;i++){if(dp[i]>=k){ans=Math.min(ans,del[i]);}}
     if(ans==Max){out.println(-1);}
     else out.println(ans);
     return;
   }
  
  //-----------------Utility--------------------------------------------
  
  static int gcd(int a,int b){if(b==0)return a; return gcd(b,a%b);}
   
  static int Max=Integer.MAX_VALUE; static long mod=1000000007;
   
  static int v(char c){return (int)(c-'a')+1;}
  
  public static long power(long x, long y )
      {
          //0^0 = 1
          long res = 1L;
          x = x%mod;
          while(y > 0)
          {
              if((y&1)==1)
                  res = (res*x)%mod;
              y >>= 1;
              x = (x*x)%mod;
          }
          return res;
      }
   
  static class Pair implements Comparable<Pair>{
          int id;int value;Pair next;
          public Pair(int id,int value) {
           
            this.id=id;this.value=value;next=null;
          }
          @Override
          public int compareTo(Pair p){return Long.compare(value,p.value);}
      }
  
  //----------------------I/O---------------------------------------------
   
  static InputStream inputStream = System.in;
  static OutputStream outputStream = System.out;
  static FastReader in=new FastReader(inputStream);
  static PrintWriter out=new PrintWriter(outputStream);
   
  static class FastReader 
  { 
    BufferedReader br; 
    StringTokenizer st; 
   
          FastReader(InputStream is) {
              br = new BufferedReader(new InputStreamReader(is));
          }
   
    public String next() 
    { 
        while (st == null || !st.hasMoreElements()) 
        { 
            try
            { 
                st = new StringTokenizer(br.readLine()); 
            } 
            catch (IOException  e) 
            { 
                e.printStackTrace(); 
            } 
        } 
        return st.nextToken(); 
    } 
   
    public int nextInt() 
    { 
        return Integer.parseInt(next()); 
    } 
   
    public long nextLong() 
    { 
        return Long.parseLong(next()); 
    } 
   
   public  double nextDouble() 
    { 
        return Double.parseDouble(next()); 
    } 
   
   
    String nextLine() 
    { 
        String str = ""; 
        try
        { 
            str = br.readLine(); 
        } 
        catch (IOException e) 
        { 
            e.printStackTrace(); 
        } 
        return str; 
    } 
  }
  static int ni() {
          return in.nextInt();
      }
  static long nl(){return in.nextLong();}
  static String ns(){return in.nextLine();}
  //---------------------I/O------------------------------------------------- 
  }