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)

#R…LやL…Rを選んでも意味がない
#L…L、R…Rを選んで反転（R…R、L…Lに）
#RL…LRを反転させると2増える
#LR…RLを反転させると2増える
#RLとLRとをそれぞれ数え上げ、min(RL数,LR数,K)を取ればよい
#例外としてLRorRLと端っこを反転したとき、1増える
#abs(LR-RL)=1 & K-min(RL,LR)>=1なら+1
#abs(LR-RL)>=1 & K-min(RL,LR)=1なら+1
#abs(LR-RL)>1 & K-min(RL,LR)>1なら+2
N,K = map(int,input().split())
S = input()
LL = 0
RR = 0
LR = 0
RL = 0
for i in range(N-1):
    if S[i]=="L" and S[i+1]=="L":
        LL += 1
    if S[i]=="R" and S[i+1]=="R":
        RR += 1
    if S[i]=="L" and S[i+1]=="R":
        LR += 1
    if S[i]=="R" and S[i+1]=="L":
        RL += 1
ans = 0
Kaisu = min(LR,RL,K)
ans += LL+RR+Kaisu*2

if abs(LR-RL)==1 and K-min(RL,LR)>=1:
    ans += 1
elif abs(LR-RL)>1 and K-min(RL,LR)==1:
    ans += 1
elif abs(LR-RL)>1 and K-min(RL,LR)>1:
    ans += 2

print(ans)