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 E1551 {
  	static int[] arr;
  	static int toBeRemoved;
  
  	static int dp(int idx, int idxInSeq) {
  		if (idxInSeq == 0) {
  			return arr[idx] == 0 ? 1 : 0;
  		}
  		int ans = -(int) 1e9;
  		for (int i = 0; i < idx; i++) {
  			ans = Math.max(ans, dp(i, idxInSeq - 1));
  		}
  		if (idxInSeq == arr[idx]) {
  			ans++;
  		}
  		return ans;
  	}
  
  	public static void main(String[] args) throws IOException {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int k = sc.nextInt();
  			arr = sc.nextIntArr(n);
  			for (int i = 0; i < n; i++)
  				arr[i]--;
  			int[][] dp = new int[n][n];
  			int[] max = new int[n];
  			for (int idxInSeq = 0; idxInSeq < n; idxInSeq++) {
  				for (int idx = 0; idx < n; idx++) {
  					if (idxInSeq == 0) {
  						dp[idxInSeq][idx] = arr[idx] == 0 ? 1 : 0;
  					} else {
  						dp[idxInSeq][idx] = (idx == 0 ? -(int) 1e9 : max[idx - 1]) + (idxInSeq == arr[idx] ? 1 : 0);
  					}
  				}
  				max = new int[n];
  				Arrays.fill(max, -(int) 1e9);
  				max[0] = dp[idxInSeq][0];
  				for (int i = 1; i < n; i++) {
  					max[i] = Math.max(max[i - 1], dp[idxInSeq][i]);
  				}
  			}
  			int ans = -1;
  			for (int i = 0; i < n; i++) {
  				if (dp[i][n - 1] >= k) {
  					ans = Math.max(ans, i + 1);
  				}
  			}
  			if (ans == -1)
  				ans = n + 1;
  			pw.println(n - ans);
  		}
  		pw.close();
  	}
  
  	static class Scanner {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public Scanner(InputStream s) {
  			br = new BufferedReader(new InputStreamReader(s));
  		}
  
  		public Scanner(FileReader f) {
  			br = new BufferedReader(f);
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public int[] nextIntArr(int n) throws IOException {
  			int[] arr = new int[n];
  			for (int i = 0; i < n; i++) {
  				arr[i] = Integer.parseInt(next());
  			}
  			return arr;
  		}
  
  	}
  
  }
  