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) {
  		InputStream inputStream = System.in;
  		OutputStream outputStream = System.out;
  		InputReader in = new InputReader(inputStream);
  		PrintWriter out = new PrintWriter(outputStream);
  		Task solver = new Task();
  		int tcCnt = in.nextInt();
  		for (int tc = 1; tc <= tcCnt; tc++)
  			solver.solve(tc, in, out);
  		out.close();
  	}
  
  	static class Task {
  		public void solve(int testNumber, InputReader in, PrintWriter out) {
  			int n = in.nextInt();
  			int k = in.nextInt();
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++) {
  				a[i] = in.nextInt();
  			}
  			final int INF = (int)(1e9);
  			// dp[i][j] is the max number of good elements after doing j moves on the first i elements
  			int[][] dp = new int[n+1][n+1];
  			for (int i = 0; i <= n; i++) {
  				Arrays.fill(dp[i], -INF);
  			}
  			dp[0][0] = 0;
  			for (int i = 1; i <= n; i++) {
  				for (int j = 0; j <= i; j++) {
  					dp[i][j] = Math.max(j > 0 ? dp[i-1][j-1] : -INF, dp[i-1][j] + (i - a[i-1] == j ? 1 : 0));
  				}
  			}
  			for (int i = 0; i <= n; i++) {
  				if (dp[n][i] >= k) {
  					out.println(i);
  					return;
  				}
  			}
  			out.println(-1);
  		}
  
  	}
  
  	static class InputReader {
  		public BufferedReader reader;
  		public StringTokenizer tokenizer;
  
  		public InputReader(InputStream stream) {
  			reader = new BufferedReader(new InputStreamReader(stream), 32768);
  			tokenizer = null;
  		}
  
  		public String nextLine() {
  			try {
  				return reader.readLine();
  			} catch (IOException e) {
  				throw new RuntimeException(e);
  			}
  		}
  
  		public String nextToken() {
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				tokenizer = new StringTokenizer(nextLine());
  			}
  			return tokenizer.nextToken();
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
  
  		public double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  
  		public long nextLong() {
  			return Long.parseLong(nextToken());
  		}
  
  		public int[] nextIntArray(int size) {
  			int[] arr = new int[size];
  			for (int i = 0; i < size; i++) {
  				arr[i] = nextInt();
  			}
  			return arr;
  		}
  
  	}
  }