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: //Created by Aminul on 7/23/2021.
  
  import java.io.*;
  import java.lang.reflect.Array;
  import java.util.*;
  
  import static java.lang.Math.*;
  
  public class E {
      static int n;
      static int[] arr;
      static int[][] dp;
  
      public static void main(String[] args) throws Exception {
          Scanner in = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
          int test = in.nextInt();
          for (int t = 1; t <= test; t++) {
              n = in.nextInt();
              int k = in.nextInt();
              arr = new int[n + 1];
              for (int i = 1; i <= n; i++) {
                  arr[i] = in.nextInt();
              }
              int res = binarySearch(0, n, k);
              pw.println(res);
          }
  
          pw.close();
      }
  
      static int binarySearch(int low, int high, int target) {
          int res = -1;
          while (low <= high) {
              int mid = low + ((high - low) >> 1);
              dp = new int[n + 1][mid + 1];
              for (int[] a : dp) {
                  Arrays.fill(a, -1);
              }
              int matches = countMatches(1, 0, mid);
              if (matches >= target) {
                  res = mid;
                  high = mid - 1;
              } else {
                  low = mid + 1;
              }
          }
          return res;
      }
  
      static int countMatches(int index, int deleted, int maxDelete) {
          if (index > n) {
              return 0;
          }
          if (dp[index][deleted] != -1) {
              return dp[index][deleted];
          }
          int isMatch = (arr[index] == index - deleted) ? 1 : 0;
          int res = isMatch + countMatches(index + 1, deleted, maxDelete);
          if (deleted < maxDelete) {
              res = max(res, countMatches(index + 1, deleted + 1, maxDelete));
          }
          return dp[index][deleted] = res;
      }
  
      static void debug(Object... obj) {
          System.err.println(Arrays.deepToString(obj));
      }
  }