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 _1551_E {
  	public static void main(String[] args) throws IOException {
  		Reader in = new Reader();
  		PrintWriter out = new PrintWriter(System.out);
  		int t = in.nextInt();
  		while(t-- > 0) {
              int n = in.nextInt();
              int k = in.nextInt();
              int[] a = new int[n];
              for(int i = 0; i < n; i++) {
                  a[i] = in.nextInt();
              }
              int max = -1;
              //dp[i][j] stores max fixed points using j elements from 0 to i - 1
              int[][] dp = new int[n + 1][n + 1];
              for(int i = 1; i <= n; i++) {
                  for(int j = 1; j <= i; j++) {
                      if(i - 1 >= j) {
                          dp[i][j] = dp[i - 1][j];
                      }
                      dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - 1] + (a[i - 1] == j ? 1 : 0));
                      if(dp[i][j] >= k) {
                          max = Math.max(max, j);
                      }
                  }
              }
              if(max == -1) {
                  out.println(max);
              }else {
                  out.println(n - max);
              }
  		}
  		in.close();
  		out.close();
  	}
  	static class Reader {
  		final private int BUFFER_SIZE = 1 << 16;
  		private DataInputStream din;
  		private byte[] buffer;
  		private int bufferPointer, bytesRead;
  		public Reader() {
  			din = new DataInputStream(System.in);
  			buffer = new byte[BUFFER_SIZE];
  			bufferPointer = bytesRead = 0;
  		}
  		public Reader(String file_name) throws IOException {
  			din = new DataInputStream(new FileInputStream(file_name));
  			buffer = new byte[BUFFER_SIZE];
  			bufferPointer = bytesRead = 0;
  		}
  		public String readLine() throws IOException {
  			byte[] buf = new byte[64]; // line length
  			int cnt = 0, c;
  			while((c = read()) != -1) {
  				if(c == '\n') {
  					if(cnt != 0) {
  						break;
  					}else {
  						continue;
  					}
  				}
  				buf[cnt++] = (byte) c;
  			}
  			return new String(buf, 0, cnt);
  		}
  		public int nextInt() throws IOException {
  			int ret = 0;
  			byte c = read();
  			while(c <= ' ') {
  				c = read();
  			}
  			boolean neg = (c == '-');
  			if(neg) {
  				c = read();
  			}
  			do{
  				ret = ret * 10 + c - '0';
  			}while((c = read()) >= '0' && c <= '9');
  			if(neg) {
  				return -ret;
  			}
  			return ret;
  		}
  		public long nextLong() throws IOException {
  			long ret = 0;
  			byte c = read();
  			while(c <= ' ') {
  				c = read();
  			}
  			boolean neg = (c == '-');
  			if(neg) {
  				c = read();
  			}
  			do{
  				ret = ret * 10 + c - '0';
  			}while((c = read()) >= '0' && c <= '9');
  			if(neg) {
  				return -ret;
  			}
  			return ret;
  		}
  		public double nextDouble() throws IOException {
  			double ret = 0, div = 1;
  			byte c = read();
  			while(c <= ' ') {
  				c = read();
  			}
  			boolean neg = (c == '-');
  			if(neg) {
  				c = read();
  			}
  			do{
  				ret = ret * 10 + c - '0';
  			}while((c = read()) >= '0' && c <= '9');
  			if(c == '.') {
  				while((c = read()) >= '0' && c <= '9') {
  					ret += (c - '0') / (div *= 10);
  				}
  			}
  			if(neg) {
  				return -ret;
  			}
  			return ret;
  		}
  		private void fillBuffer() throws IOException {
  			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
  			if(bytesRead == -1) {
  				buffer[0] = -1;
  			}
  		}
  		private byte read() throws IOException {
  			if(bufferPointer == bytesRead) {
  				fillBuffer();
  			}
  			return buffer[bufferPointer++];
  		}
  		public void close() throws IOException {
  			if(din == null) {
  				return;
  			}
  			din.close();
  		}
  	}
  }
  