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.io.*;
  import java.util.*;
  public class Main {
  	public static void main(String[] args) throws IOException 
  	{ 
  		FastScanner f = new FastScanner(); 
  		int ttt=1;
  		ttt=f.nextInt();
  		PrintWriter out=new PrintWriter(System.out);
  		outer:for(int tt=0;tt<ttt;tt++) {
  			int n=f.nextInt(); int k=f.nextInt();
  			int[] l=f.readArray(n);
  			int[][] dp=new int[n+1][n+1];
  			for(int i=0;i<=n;i++){
  		        for(int j=0;j<=n;j++) {
  		            dp[i][j] = -1;
  		        }
  		    }
  		    dp[0][0]= 0;
  		    for(int i=0;i<n;i++) {
  		        for(int j=0;j<=i;j++) {
  		            // delete
  		            dp[i+1][j+1] =Math.max(dp[i+1][j+1], dp[i][j]);
  		            // don't delete
  		            dp[i+1][j] = Math.max(dp[i+1][j], dp[i][j] + (l[i]== i - j+1 ? 1 : 0));
  		        }
  		    }
  			for(int i=0;i<=n;i++){
  				if(dp[n][i] >= k){
  					System.out.println(i);
  					continue outer;
  				}
  			}
  			
  			System.out.println(-1);
  		}
  		out.close();
  	} 
  	static void sort(int[] p) {
          ArrayList<Integer> q = new ArrayList<>();
          for (int i: p) q.add( i);
          Collections.sort(q);
          for (int i = 0; i < p.length; i++) p[i] = q.get(i);
      }
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		long[] readLongArray(int n) {
  			long[] a=new long[n];
  			for (int i=0; i<n; i++) a[i]=nextLong();
  			return a;
  		}
  	}
  }