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 sys, os
  from io import BytesIO, IOBase
  from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
  from collections import defaultdict as dd, deque
  from heapq import merge, heapify, heappop, heappush, nsmallest
  from bisect import bisect_left as bl, bisect_right as br, bisect
  
  # region fastio
  BUFSIZE = 8192
   
  class FastIO(IOBase):
      newlines = 0
   
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
   
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
   
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
   
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
   
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  stdin, stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) 
  
  mod = pow(10, 9) + 7
  mod2 = 998244353
          
  def inp(): return stdin.readline().strip()
  def iinp(): return int(inp())
  def out(var, end="\n"): stdout.write(str(var)+"\n")
  def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
  def lmp(): return list(mp())
  def mp(): return map(int, inp().split())
  def l1d(n, val=0): return [val for i in range(n)]
  def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
  def ceil(a, b): return (a+b-1)//b
  S1 = 'abcdefghijklmnopqrstuvwxyz'
  S2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
  def isprime(x):
      if x<=1: return False
      if x in (2, 3): return True
      if x%2 == 0: return False
      for i in range(3, int(sqrt(x))+1, 2):
          if x%i == 0: return False
      return True
  
  for _ in range(int(inp())): 
      n, k = mp()
      arr = lmp()
      dp = l2d(n, n)
      ans = inf
      for i in range(n):
          for j in range(i+1):
              if arr[i] == j+1:
                  dp[i][j] = 1+dp[i-1][j-1]
              else:
                  dp[i][j] = max(dp[i-1][j], dp[i-1][j-1])
              if dp[i][j] == k:
                  ans = min(ans, i-j)
      print(ans if ans != inf else -1)