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.*;
import java.math.*;

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 result = 0;
        char prev = S.charAt(0);
        char firstC = prev;
        int leftIndexTmp = 1;
        boolean leftSame = true;
        
        for(int i = 1; i < N; i++){
            char now = S.charAt(i);
            if(leftSame && firstC == now){
                leftIndexTmp++;
            }else{
                leftSame = false;
            }
            if(prev == now){
                if(prev == 'R'){
                    result++;
                }
                if(now == 'L'){
                    result++;
                }
            }
            prev = now;
        }
        int leftIndex = 0;
        int rightIndex = N - 1;
        if(S.charAt(0) == S.charAt(N - 1)){
            char targetC = S.charAt(0);
            while(S.charAt(leftIndex) == targetC){
                leftIndex++;
            }
            while(S.charAt(rightIndex) == targetC){
                rightIndex--;
            }
        }else{
            rightIndex = N - 2;
            char lastC = S.charAt(N - 1);
            for(int i = N - 2; i >= 0; i--){
                if(S.charAt(i) == lastC){
                    rightIndex--;
                }else{
                    break;
                }
            }
            leftIndex = leftIndexTmp > (N - rightIndex - 1) ? leftIndexTmp : 0;
            rightIndex = leftIndexTmp > (N - rightIndex - 1) ? N - 1 : rightIndex;
        }
        int nowC = S.charAt(leftIndex);
        int count = 0;
        while(count < K){
            while(S.charAt(leftIndex) != nowC){
                leftIndex++;
            }
            while(S.charAt(rightIndex) != nowC){
                rightIndex--;
            }
            if(leftIndex < rightIndex){
                count++;
            }else{
                break;
            }
            nowC = nowC == 'R' ? 'L' : 'R';
        }
        // 最後に残った区間が全て同じ文字であれば -1
        char searchC = S.charAt(leftIndex);
        int minus = 1;
        for(int i = leftIndex; i <= rightIndex; i++){
            if(searchC != S.charAt(i)){
                minus = 0;
                break;
            }
        }
        System.out.println(result + (count * 2) - minus);
    }
}