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: from collections import deque
  import heapq
  from math import sqrt
  import operator
  from operator import itemgetter
  import sys
  
  import copy
  
   
  inf_var = 0
  if inf_var == 1:
      inf = open('input.txt', 'r')
  else:
      inf = sys.stdin
   
  input = inf.readline
  # sys.stdout = open('output.txt', 'w')
   
  def read_one_int():
      return int(input().rstrip('\n'))
  
  
  def read_one_float():
      return float(input().rstrip('\n'))
   
   
  def read_list_of_ints():
      res = [int(val) for val in (input().rstrip('\n')).split(' ')]
      return res
   
  def read_str():
      return input().rstrip()
  
  
  def main():
      tests = read_one_int()
  
      for _ in range(tests):
          n, k = read_list_of_ints()
          elements = read_list_of_ints()
  
          graph = [[] for _ in range(n)]
  
          for i in range(n - 1, -1, -1):
              need_to_del_i = i + 1 - elements[i]
              if need_to_del_i < 0:
                  continue
              for j in range(i - 1, -1, -1):
                  need_to_del_j = j + 1 - elements[j]
  
                  if need_to_del_j < 0 or need_to_del_j > need_to_del_i:
                      continue
  
                  if j < need_to_del_j or need_to_del_i - need_to_del_j > i - j - 1:
                      continue
  
                  graph[i].append(j)
  
          # print(graph)
          cache = {}
  
          def count_max(v):
              if v in cache:
                  return cache[v]
  
              if v + 1 - elements[v] >= 0:
                  res = 0
                  for u in graph[v]:
                      res = max(res, count_max(u))
  
                  cache[v] = res + 1
                  return res + 1
              else:
                  cache[v] = 0
                  return 0            
  
          res = -1
          for i in range(n):
              cur_res = count_max(i)
  
              if cur_res >= k:
                  if res == -1:
                      res = i + 1 - elements[i]
                  else:
                      res = min(res, i + 1 - elements[i])
  
          print(res)
  
  
  if __name__== '__main__':
      main()
  