Problem:
Score : 400 points

Problem 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?

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

|S| = N

Each character of S is L or R.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of happy people after at most K operations.

Sample Input 1

6 1
LRLRRL

Sample Output 1

3

If we choose (l, r) = (2, 5), we have LLLRLL, where the 2-nd, 3-rd, and 6-th persons from the west are happy.

Sample Input 2

13 3
LRRLRLRRLRLLR

Sample Output 2

9

Sample Input 3

10 1
LLLLLRRRRR

Sample Output 3

9

Sample Input 4

9 2
RRRLRLRLL

Sample Output 4

7

Solution in Python (3.4.3) (judge verdict: Accepted)

#!/usr/bin/env python3
import sys
import math
import numpy as np
# from functools import lru_cache

args = sys.argv
if len(args) == 2:
  from IPython import embed
  em = True
else:
  em = False

#### input
N, K = map(int, input().split())
S = input()
# a = list(map(int, input().split()))
# a = list(map(lambda x: int(x)*(-1), a))  # 各要素を-1倍


# n = int(input())
# t = [int(input()) for i in range(n)]
if em: sys.stdin = open('/dev/tty')

#### functions
# @lru_cache(maxsize=10000)

#### process
# if em: embed()

score = 0
for i in range(0, len(S) - 1):
  if(S[i] == S[i+1]): score = score + 1

num_R = len([s for s in S if s == "R"])

if num_R > len(S) / 2:
  S2 = S.split("R")
else:
  S2 = S.split("L")

edge     = len([s for s in [S2[0],S2[-1]] if s != ""])
non_edge = len([s for s in S2[1:-1] if s != ""])

# if em:embed()

score = score + 2 * min(K, non_edge)
K = K - min(K, non_edge)
score = score + min(K, edge)

print(score)

## heapsort, O(logN)
# import heapq
# heapq.heapify(A)
# min = heapq.heappop(A)
# heapq.heappush(A, new_element)