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 (
	"fmt"
	"sort"
)

func main() {
	var n, k int
	var s string
	fmt.Scan(&n, &k)
	fmt.Scanf("%s", &s)
	lastChar := s[0]
	count := 0
	dups := make([]int, 0)
	for i := 1; i < n; i++ {
		if s[i] == lastChar {
			count++
		} else {
			dups = append(dups, count+1)
			count = 0
			lastChar = s[i]
		}
	}
	dups = append(dups, count+1)
	sort.Ints(dups)
	happyPersonNum := 0
	c := 0
	if len(dups)-1 <= k {
		happyPersonNum = n - 1
	} else {
		for i := len(dups) - 1; i >= 0; i-- {
			happyPersonNum += dups[i] + 1
			c++
			if c >= k {
				break
			}
		}
	}
	fmt.Println(happyPersonNum)
}