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: Wrong Answer)

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

if n == 1:
        print(0)
else:        
        t = []
        tt = 0
        ss = s[0]
        for i in range(1,n):
                if s[i] != ss:
                        t.append(i - tt)
                        tt = i
                        ss = s[i]
        t.append(n - tt)
        cnt = sum(t) - len(t) 

        if len(t) % 2 == 0:
                if k <= (len(t) - 2)/2:
                        cnt += 2 * k
                else:
                        cnt += 2 * (len(t) - 2)/2 + 1
        else:
                if k <= (len(t) - 1)/2:
                        cnt += 2 * k
                elif k == (len(t) - 1)/2 + 1:
                        cnt += 2 * k + 1
                else:
                        cnt += 2 *(len(t) - 1)/2 + 2
        print(int(cnt))