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()
  s = list(s)
  print(s)
  before = s[1]
  flag = False
  place = -1
  end = -1
  cnt = 0
  
  for i in range(1, n - 1):
  
      if s[i] != before:
          flag = True
          place = i
  
      if flag:
          print(s, i)
          if i + 1 == n - 1:
              if s[place] == s[i + 1]:
                  end = i + 1
              else:
                  end = i
              print(s, i, end, "end")
              if s[place] == "L":
                  if end == i:
                      s[place:end] = "R"
                  else:
                      s[place:end] = "R"
                      # s[end-1] ='R'
              else:
                  if end == i:
                      s[place:end] = "L"
                  else:
                      s[place:end] = "L"
                      # s[end-1] = 'L'
              print(s, "end2")
              cnt += 1
              if cnt == k:
                  break
              flag = False
              before = s[i]
              continue
  
          if s[i] != s[i + 1]:
              print(s)
              if s[place] == "L":
                  s[place : i + 1] = "R"
              else:
                  s[place : i + 1] = "L"
  
              cnt += 1
              print(s)
              if cnt == k:
                  break
              flag = False
              before = s[i]
              continue
      before = s[i]
  
  print(s)
  if cnt != k:
      if s[0] != s[1]:
          s[0] = s[1]
      if k >= cnt + 2:
          if s[n - 1] != s[n - 2]:
              s[n - 1] = s[n - 2]
  
  seqCnt = 0
  ans = 0
  print(s)
  for i in range(n - 1):
      print(i, seqCnt)
      if s[i] == s[i + 1]:
          seqCnt += 1
      if s[i] != s[i + 1] or i == n - 2:
          print(ans, seqCnt, i)
          if seqCnt >= 1:
              ans += seqCnt
          seqCnt = 0
  
  print(ans)
  
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"}]