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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<string>

int main()
{
    int N,K;
    std::string S;
    std::cin>>N>>K;
    std::cin>>S;
    int p=0;
    int n=0;
    char past = S[0];
    //pの数を数える
    for(int i=1;i<N;i++){
        if(past-6==S[i])p++;
        past = S[i];
    }
    n = p*2;
    if(S[0]=='L')n++;
    if(S[N-1]=='R')n++;

    if(p==0&&S[0]=='L'&&S[N-1]=='R'){std::cout<<N-1<<std::endl;return 0;}
    if(p>=K-2){
        n -= K*2;
    }else{
        for(int i=0;i<K||n==0||n==1;i++)n -= 2;
        if(S[0]=='L'&&S[N-1]=='R')n++;
        //std::cout<<<<std::endl;
    }
    std::cout<<N-n<<std::endl;

    return 0;
}