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: Runtime Error)

# -*- coding: utf-8 -*-
N, K = map(int, input().split())
S = input()
out = ['0'] * len(S)
for i in range(K):
    if i!=0:
        S = ''.join(out)
    idx = 0
    maxidx = 0
    length = 1
    maxlen = 10000
    back = S[0]
    for j in range(len(S)):
        if back == S[j]:
            length += 1
        else:
            if length <= maxlen:
                #print(length)
                maxlen = length
                maxidx = j
                length = 1
            else:
                length = 1
                idx = j
        if length == j+2 and S[0] == 'L':
            length -= 1
        if j == len(S)-1 and S[j] == 'R':
            length -= 1
        back = S[j]
    temp = ['0'] * maxlen
    for j in range(maxlen):
        #print([maxidx,maxlen])
        if maxlen == 1:
            if S[maxidx + maxlen - j - 2] == 'L':
                temp[j] = 'R'
            else:
                temp[j] = 'L'
        else:
            if S[maxidx + maxlen - j - 1] == 'L':
                temp[j] = 'R'
            else:
                temp[j] = 'L'
    for j in range(len(S)):
        out[j] = S[j]
    for j in range(maxlen):
        if maxlen == 1:
            out[maxidx + j - 1] = temp[j]
        else:
            out[maxidx + j] = temp[j]
#print(''.join(out))
count=0
#right
for i in range(len(S)):
    if i==0 and out[i]=='L':
        continue
    elif i==N-1 and out[i]=='R':
        break
    if out[i]=='R' and out[i+1]=='R':
        count += 1
    if out[i]=='L' and out[i-1]=='L':
        count += 1
print(count)