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)

def count_happiness(S):
    happiness = 0
    for n in range(N):
        if(S[n] == "L" and n > 0 and S[n] == S[n-1]):
            happiness += 1
        elif(S[n] == "R" and n < N -1 and S[n] == S[n+1]):
            happiness += 1
    return happiness

N, K = map(int, input().split())
S = list(input())

LR_RL_unhappy_pair = 0
LR_unhappy = 0
RL_unhappy = 0
for n in range(N-1):
    if(S[n] == "R" and S[n+1] == "L"):
        RL_unhappy += 1
    if(S[n] == "L" and S[n+1] == "R"):
        LR_unhappy += 1

change_pair_count  = min(K, min(RL_unhappy, LR_unhappy))
change_single_count = 0
if(K > change_pair_count):
    remain = K - change_pair_count
    changable = abs(RL_unhappy  - LR_unhappy)
    change_single_count = min(RL_unhappy, changable) 
print(count_happiness(S) + 2 * change_pair_count + change_single_count)