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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>

unsigned int func(unsigned int n, int* a){
    unsigned int count = 0;
    for(int i=1; i<=n; i++){
        if(a[i] == a[i + a[i]]){
            count++;
        }
    }
    return count;
}

int main(){
    unsigned int n, k;
    scanf("%u %u", &n, &k);
    char s[n+1];
    scanf("%s", s);
    int a[n+200];
    a[0] = 1;
    a[n+1] = -1;
    for(int i=0; i<n; i++){
        if(s[i] == 'L'){
            a[i+1] = -1;
        }else{
            a[i+1] = 1;
        }
    }
    for(unsigned int i=0; i<k; i++){
        unsigned int t, u = -1;
        for(unsigned int j=2; j<n; j++){
            if(a[j-1] != a[j]){
                t = j;
                for(unsigned int w=j; w<n+1; w++){
                    u=w;
                    if(a[t-1] == a[w+1]){
                        break;
                    }
                }
                break;
            }
        }
        //printf("%lu %lu\n", t, u);
        for(unsigned int j=t; j<=u; j++){
            a[j] *= -1;
        }
    }
    /*for(int i=1; i<n+1; i++){
        printf("%d ", a[i]);
    }
    printf("\n");*/
    printf("%u", func(n, a));

    return 0;
}