output_description: Print the maximum possible number of happy people after at most K operations.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N K
      S
attempt: N, K = map(int, input().split())
  S = input()
  
  same_L_cnt = 0
  
  up_cnt = 0
  same_cnt = 0
  before = ""
  buf = []
  
  
  for i in range(N):
      now_S = S[i]
      if now_S == "L":
          now_S = 1
      else:
          now_S = 0
      # 前回と同じか確認
      if now_S == before:
          # 同じならカウントアップ
          same_cnt += 1
      # L R R ..... R Lが出現したか確認する
      buf.append(now_S)
      # print('------ buffer ------')
      # pprint(buf)
      # print('--------------------')
      if buf[0] != 1:
          buf = []
      elif len(buf) == 2 and buf[1] == 1:
          buf = [1]
      elif len(buf) < 3:
          pass
      else:
          t = buf[1 : len(buf) - 2]
          if sum(t) != 0:
              buf = []
          elif buf[len(buf) - 1] == 1:
              up_cnt += 1
              buf = [1]
          else:
              pass
      before = now_S
  # print(up_cnt)
  # print(same_cnt)
  up_flg = 0
  if sum(buf) == 1:
      up_flg = 1
  
  pt = same_cnt
  rev_cnt = min(K, up_cnt)
  remain_cnt = K - rev_cnt
  extra = 0
  if up_flg and remain_cnt > 0:
      extra = 1
  print(same_cnt + rev_cnt * 2 + extra)
  
problem_description: Statement
  
  There are N people standing in a queue from west to east.
  
  Given is a string S of length N representing the directions of the people. The
  i-th person from the west is facing west if the i-th character of S is `L`,
  and east if that character of S is `R`.
  
  A person is happy if the person in front of him/her is facing the same
  direction. If no person is standing in front of a person, however, he/she is
  not happy.
  
  You can perform the following operation any number of times between 0 and K
  (inclusive):
  
  Operation: Choose integers l and r such that 1 \leq l \leq r \leq N, and
  rotate by 180 degrees the part of the queue: the l-th, (l+1)-th, ..., r-th
  persons. That is, for each i = 0, 1, ..., r-l, the (l + i)-th person from the
  west will stand the (r - i)-th from the west after the operation, facing east
  if he/she is facing west now, and vice versa.
  
  What is the maximum possible number of happy people you can have?
samples: [{"input": "6 1\n    LRLRRL", "output": "3\n    \n\nIf we choose (l, r) = (2, 5), we have `LLLRLL`, where the 2-nd, 3-rd, and 6-th\npersons from the west are happy.\n\n* * *"}, {"input": "13 3\n    LRRLRLRRLRLLR", "output": "9\n    \n\n* * *"}, {"input": "10 1\n    LLLLLRRRRR", "output": "9\n    \n\n* * *"}, {"input": "9 2\n    RRRLRLRLL", "output": "7"}]