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.ByteArrayInputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.Arrays;
  import java.util.InputMismatchException;
  
  public class Main {
  
  	private static final String NO = "NO";
  	private static final String YES = "YES";
  	InputStream is;
  	PrintWriter out;
  	String INPUT = "";
  
  	private static final long MOD = (long) (1e9 + 7);
  
  	void solve() {
  		int T = ni();
  		for (int i = 0; i < T; i++)
  			solve(i);
  	}
  
  	void solve(int nth) {
  		int n = ni();
  		int k = ni();
  		int a[] = na(n);
  		int c[] = new int[n];
  		for (int i = 0; i < n; i++)
  			c[i] = i + 1 - a[i];
  		int dp[] = new int[n];
  		int ans = Integer.MAX_VALUE;
  		for (int i = 0; i < n; i++) {
  			if (c[i] >= 0) {
  				dp[i] = 1;
  				for (int j = 0; j < i; j++) {
  					if (a[i] > a[j] && c[i] >= c[j])
  						dp[i] = Math.max(dp[i], dp[j] + 1);
  				}
  				if (dp[i] >= k)
  					ans = Math.min(ans, c[i]);
  			}
  		}
  		out.println(ans == Integer.MAX_VALUE ? -1 : ans);
  	}
  
  	void run() throws Exception {
  		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
  		out = new PrintWriter(System.out);
  
  		long s = System.currentTimeMillis();
  		solve();
  		out.flush();
  		if (!INPUT.isEmpty())
  			tr(System.currentTimeMillis() - s + "ms");
  	}
  
  	public static void main(String[] args) throws Exception {
  		new Main().run();
  	}
  
  	private byte[] inbuf = new byte[1024];
  	public int lenbuf = 0, ptrbuf = 0;
  
  	private int readByte() {
  		if (lenbuf == -1)
  			throw new InputMismatchException();
  		if (ptrbuf >= lenbuf) {
  			ptrbuf = 0;
  			try {
  				lenbuf = is.read(inbuf);
  			} catch (IOException e) {
  				throw new InputMismatchException();
  			}
  			if (lenbuf <= 0)
  				return -1;
  		}
  		return inbuf[ptrbuf++];
  	}
  
  	private boolean isSpaceChar(int c) {
  		return !(c >= 33 && c <= 126);
  	}
  
  	private int skip() {
  		int b;
  		while ((b = readByte()) != -1 && isSpaceChar(b))
  			;
  		return b;
  	}
  
  	private double nd() {
  		return Double.parseDouble(ns());
  	}
  
  	private char nc() {
  		return (char) skip();
  	}
  
  	private char[] nc(int n) {
  		char[] ret = new char[n];
  		for (int i = 0; i < n; i++)
  			ret[i] = nc();
  		return ret;
  	}
  
  	private String ns() {
  		int b = skip();
  		StringBuilder sb = new StringBuilder();
  		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != '
  									// ')
  			sb.appendCodePoint(b);
  			b = readByte();
  		}
  		return sb.toString();
  	}
  
  	private char[] ns(int n) {
  		char[] buf = new char[n];
  		int b = skip(), p = 0;
  		while (p < n) {
  			if (!(isSpaceChar(b)))
  				buf[p++] = (char) b;
  			b = readByte();
  		}
  		return n == p ? buf : Arrays.copyOf(buf, p);
  	}
  
  	private char[][] nm(int n, int m) {
  		char[][] map = new char[n][];
  		for (int i = 0; i < n; i++)
  			map[i] = ns(m);
  		return map;
  	}
  
  	private int[] na(int n) {
  		int[] a = new int[n];
  		for (int i = 0; i < n; i++)
  			a[i] = ni();
  		return a;
  	}
  
  	private Integer[] na2(int n) {
  		Integer[] a = new Integer[n];
  		for (int i = 0; i < n; i++)
  			a[i] = ni();
  		return a;
  	}
  
  	private int[][] na(int n, int m) {
  		int[][] a = new int[n][];
  		for (int i = 0; i < n; i++)
  			a[i] = na(m);
  		return a;
  	}
  
  	private int ni() {
  		int num = 0, b;
  		boolean minus = false;
  		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
  			;
  		if (b == '-') {
  			minus = true;
  			b = readByte();
  		}
  
  		while (true) {
  			if (b >= '0' && b <= '9') {
  				num = num * 10 + (b - '0');
  			} else {
  				return minus ? -num : num;
  			}
  			b = readByte();
  		}
  	}
  
  	private Long[] nl2(int n) {
  		Long[] a = new Long[n];
  		for (int i = 0; i < n; i++)
  			a[i] = nl();
  		return a;
  	}
  
  	private long[] nl(int n) {
  		long[] a = new long[n];
  		for (int i = 0; i < n; i++)
  			a[i] = nl();
  		return a;
  	}
  
  	private long[][] nl(int n, int m) {
  		long[][] a = new long[n][];
  		for (int i = 0; i < n; i++)
  			a[i] = nl(m);
  		return a;
  	}
  
  	private long nl() {
  		long num = 0;
  		int b;
  		boolean minus = false;
  		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
  			;
  		if (b == '-') {
  			minus = true;
  			b = readByte();
  		}
  
  		while (true) {
  			if (b >= '0' && b <= '9') {
  				num = num * 10 + (b - '0');
  			} else {
  				return minus ? -num : num;
  			}
  			b = readByte();
  		}
  	}
  
  	private static void tr(Object... o) {
  		System.out.println(Arrays.deepToString(o));
  	}
  
  }
  