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 Go (1.6) (judge verdict: Wrong Answer)

package main
 
import (
	"bufio"
	"fmt"
	"log"
	"os"
	"strconv"
)
 
func run() {
	n := nextInt()
	k := nextInt()
	s := next()
	
	v := make([]int, 0)
	
	for i := 0; i < n; i++ {
		if(string(s[i]) == "R"){
			v = append(v, 1)
		}else {
			v = append(v, 0)
		}
	}
	
	count := 0
	count2 :=0
	te := v[0]
	
	for i := 1; i < n; i++ {
		if(te == v[i]){ 
			count ++
		}else{
			count2 ++
			te = v[i]
		}
	} 
	
	if(count2 == 0 && k == 1){
		Println(count - 1)
	} else if ((count2+1 / 2) < k) {
		Println(count + (count2+1 / 2) * 2)
	} else{
		Println(count + k * 2)
	}
	
}

// --- template ---
// Thanks https://gist.github.com/halllllll/853ab587fd82ee3ffe6f7fb7fb499695
 
var sc = bufio.NewScanner(os.Stdin)
var out = bufio.NewWriter(os.Stdout)
 
// Caution: You will get unexpected result if a single token is larger than this size.
const BufferSize = 64 * 1024 * 1024
 
func Print(a ...interface{}) (n int, err error) {
	return fmt.Fprint(out, a...)
}
 
func Printf(format string, a ...interface{}) (n int, err error) {
	return fmt.Fprintf(out, format, a...)
}
 
func Println(a ...interface{}) (n int, err error) {
	return fmt.Fprintln(out, a...)
}
 
func next() string {
	sc.Scan()
	return sc.Text()
}
 
func nextInt() int {
	a, err := strconv.Atoi(next())
	if err != nil {
		log.Fatal(err)
	}
	return a
}
 
func nextFloat() float64 {
	a, err := strconv.ParseFloat(next(), 64)
	if err != nil {
		log.Fatal(err)
	}
	return a
}

 
func main() {
	sc.Buffer(make([]byte, 0, BufferSize), BufferSize)
	sc.Split(bufio.ScanWords)
	run()
	out.Flush()
}