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: Accepted)

#include <iostream>
#include <cmath>
using namespace std;

int main(){
    int ANS = 0;

    int N, K;
    cin >> N >> K;

    char temp[N];
    int stats[N];
    int LR = 0;
    int RL = 0;
    int limit;
    for(int i=0; i<N; i++){
        cin >> temp[i];
        if(temp[i] == 'L') stats[i] = 0;
        else stats[i] = 1;

        if(i!=0){
            if(stats[i-1]==stats[i]) ANS++;
            else if(stats[i-1]==0 && stats[i]==1) LR++;
            else RL++;
        }
    }

    limit = min(LR, RL);

    for(int i=0; i<K; i++){
        if(LR>0 && RL>0){
            ANS += 2;
            LR--;
            RL--;
        }else if(LR>0 || RL>0){
            if(LR>0){
                ANS++;
                LR--;
            }else{
                ANS++;
                RL--;
            }
        }else{
            break;
        }
    }

    cout << ANS << endl;
}