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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;
public class Main {
    public static void main(String args []) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        String S = sc.next();
        int [] s = new int[S.length()];
        for(int i=0;i<S.length();i++){
            if(S.charAt(i)=='R') s[i]=0;
            else s[i] =1;
        }
        int r_num=0,l_num=0;
        for(int i=0;i<S.length();i++){
            if(s[i]==0) r_num++;
            else l_num++;
        }
        int opt =0;
        if(r_num>l_num){
            for(int i=0;i<K;i++){
                int first=-1,max_first=-1,fin=0,max_fin=0,max=0,num=0;
                for(int j=0;j<N;j++){
                    if(first==-1&&s[j]==1) first=j;
                    if(s[j]==1){
                        num++;
                        fin = j;
                        if(num>=max&&j==N-1){
                            if(num>=max){
                                max = num ;
                                max_first = first;
                                max_fin = fin;
                            }
                        }
                    }
                    else{
                        if(num>=max){
                            max = num ;
                            max_first = first;
                            max_fin = fin;
                        }
                        num =0;
                        first = -1;
                        fin = 0;
                    }
                }
                for(int j=max_first;j<=max_fin;j++) s[j]=0;
            }
            int num = 0;
            for(int i=0;i<N;i++){
                if(s[i]==0&&i==N-1) opt+=num;
                else if(s[i]==0) num++;
                else{
                    if(num!=0)opt+=(num-1);
                    num=0;
                }
            }
        }
        else{
            for(int i=0;i<K;i++){
                int first=-1,max_first=-1,fin=0,max_fin=0,max=0,num=0;
                for(int j=0;j<N;j++){
                    if(first==-1&&s[j]==0) first=j;
                    if(s[j]==0){
                        num++;
                        fin = j;
                        if(num>=max&&j==N-1){
                            if(num>=max){
                                max = num ;
                                max_first = first;
                                max_fin = fin;
                            }
                        }
                    }
                    else if(s[j]==1||j==N-1){
                        if(num>=max){
                            max = num ;
                            max_first = first;
                            max_fin = fin;
                        }
                        num =0;
                        first = -1;
                        fin = 0;
                    }
                }
                for(int j=max_first;j<=max_fin;j++) s[j]=1;
            }
            int num = 0;
            for(int i=0;i<N;i++){
                if(s[i]==1&&i==N-1) opt+=(num);
                else if(s[i]==1) num++;
                else{
                    if(num!=0)opt+=(num-1);
                    num=0;
                }
            }
        }
        System.out.println(opt);
    }
}