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 defaultdict, deque, Counter
  from heapq import heapify, heappop, heappush
  import math
  from copy import deepcopy
  from itertools import combinations, permutations, product, combinations_with_replacement
  from bisect import bisect_left, bisect_right
  
  import sys
  def input():
      return sys.stdin.readline().rstrip()
  def getN():
      return int(input())
  def getNM():
      return map(int, input().split())
  def getList():
      return list(map(int, input().split()))
  def getListGraph():
      return list(map(lambda x:int(x) - 1, input().split()))
  def getArray(intn):
      return [int(input()) for i in range(intn)]
  
  mod = 10 ** 9 + 7
  MOD = 998244353
  # sys.setrecursionlimit(10000000)
  inf = float('inf')
  eps = 10 ** (-10)
  dy = [0, 1, 0, -1]
  dx = [1, 0, -1, 0]
  
  #############
  # Main Code #
  #############
  
  
  
  T = getN()
  for _ in range(T):
      N, K = getNM()
      A = [0] + getList()
  
      # 現在までにj個消した時の最大適合
      prev = [0] * (N + 1)
  
      # i個目まで進んだ
      for i in range(1, N + 1):
          next = [0] * (N + 1)
          # 前回までにi個まで消した
          for j in range(i):
              # i個目を削除する場合
              next[j + 1] = max(next[j + 1], prev[j])
              # 削除しない場合　適合するかチェック
              # 今回のはi - j個目　これがA[i]と一致するかチェック
              next[j] = max(next[j], prev[j] + ((i - j) == A[i]))
  
          prev = next
  
      for i in range(N + 1):
          if prev[i] >= K:
              print(i)
              break
      else:
          print(-1)