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: Time Limit Exceeded)

n, k = map(int, input().split())
s = input()
turn = 0

happiness = 0


def exchange(num, loc):
    global s
    tmp = s[loc:loc+num+1]
    tmp = tmp[::-1]
    tmp = ['L' if x == 'R' else 'R' for x in tmp]
    s[loc:loc+num+1] = tmp

for i in range(n):
    if i == 0:
        if s[i] == 'R' and s[i+1] == 'R':
            happiness += 1
    elif i == n-1:
        if s[i] == 'L' and s[i-1] == 'L':
            happiness += 1
    else:
        if s[i] == 'R' and s[i+1] == 'R':
            happiness += 1
        elif s[i] == 'L' and s[i-1] == 'L':
            happiness += 1

for i in range(n-1):
    if s[i] == 'R' and s[i+1] == 'L' or s[i] == 'L' and s[i+1] == 'R':
        turn += 1


s = 'X' + s + 'X'
s = list(s)
lamp = -1
while k != 0:
    lamp += 1
    if turn == 1:
        print(n-1)
        exit()
    for j in range(2, n-lamp-1):
        if s[j] == 'R' and s[j - 1] != 'R' and s[j + lamp] == 'R' and s[j + lamp + 1] != 'R':
            happiness += 2
            exchange(lamp, j)
            k -= 1
            turn -= 2
            break
        elif s[j] == 'L' and s[j - 1] != 'L' and s[j + lamp] == 'L' and s[j + lamp + 1] != 'R':
            happiness += 2
            exchange(lamp, j)
            k -= 1
            turn -= 2
            break
    if turn == 0:
        break

print(happiness)