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.util.Map.Entry;
  
  import java.math.*;
  import java.io.*;
  
  public class Main {
  	public static void main(String[] args) throws FileNotFoundException {
  		InputReader in = new InputReader(System.in);
  		// Scanner in = new Scanner(System.in);
  		// Scanner in = new Scanner(new BufferedReader(new
  		// InputStreamReader(System.in)));
  		PrintWriter out = new PrintWriter(System.out);
  		// InputReader in = new InputReader(new
  		// File("ethan_traverses_a_tree.txt"));
  		// PrintWriter out = new PrintWriter(new
  		// File("ethan_traverses_a_tree-output.txt"));
  
  		int pi = in.nextInt();
  		for (int qi = 0; qi < pi; qi++) {
  			int n = in.nextInt();
  			int m = in.nextInt();
  
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++) {
  				a[i] = in.nextInt();
  			}
  
  			int[][] dp = new int[n][n + 1];
  			for (int i = 0; i < n; i++) {
  				if (i == 0) {
  					if (a[i] == i + 1) {
  						dp[i][0] = 1;
  					} else {
  						dp[i][0] = 0;
  					}
  
  					dp[i][1] = 0;
  				} else {
  					if (a[i] == i + 1) {
  						dp[i][0] = dp[i - 1][0] + 1;
  					} else {
  						dp[i][0] = dp[i - 1][0];
  					}
  
  					for (int j = 1; j <= i; j++) {
  						int max = dp[i - 1][j - 1];
  
  						if (a[i] == i + 1 - j) {
  							max = Math.max(max, dp[i - 1][j] + 1);
  						} else {
  							max = Math.max(max, dp[i - 1][j]);
  						}
  
  						dp[i][j] = max;
  					}
  
  					dp[i][i + 1] = 0;
  				}
  			}
  
  			int ans = -1;
  			for (int j = 0; j <= n; j++) {
  				if (dp[n - 1][j] >= m) {
  					ans = j;
  					break;
  				}
  			}
  
  			out.printf("%d\n", ans);
  		}
  
  		out.close();
  	}
  
  	static class InputReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public InputReader(File f) {
  			try {
  				br = new BufferedReader(new FileReader(f));
  			} catch (FileNotFoundException e) {
  				e.printStackTrace();
  			}
  		}
  
  		public InputReader(InputStream in) {
  			br = new BufferedReader(new InputStreamReader(in));
  		}
  
  		public String next() {
  			while (st == null || !st.hasMoreTokens()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		public boolean hasNext() {
  			while (st == null || !st.hasMoreTokens()) {
  				String s = null;
  				try {
  					s = br.readLine();
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  				if (s == null)
  					return false;
  				st = new StringTokenizer(s);
  			}
  			return true;
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		public double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  }