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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <string>
#include <iostream>
#include <vector>

using namespace std;


int main()
{
    int N, K,count,STRSIZE;
    string str;
    vector<char> n_str,tmp;

    cin >> N >> K >> str;

    //cout << str << "\n";
    count = 0;

    for(int i=0; i<(int)str.size(); i++){
        n_str.push_back(str[i]);
    }

    STRSIZE = (int)n_str.size();

    if(n_str[0] != 'R' || n_str[1] != 'R') tmp.push_back(n_str[0]);
    if(n_str[0] == 'R' && n_str[1] == 'R') count++;

    for(int i = 1; i < STRSIZE -1; ++i){
        //cout << "i: " << i << "\n";
        if(n_str[i] == 'R'){
            if(n_str[i+1] == 'R'){
                count++;
                //cout << " count: "  << count << "\n";
                continue;
            }
        }
        if(n_str[i] == 'L'){
            if(n_str[i-1] == 'L'){
                count++;
                //cout << "count: "  << count << "\n";
                continue;
            }
        }
        tmp.push_back(n_str[i]);
    }

    if(STRSIZE >= 2){
        if(n_str[STRSIZE -2] != 'L' || n_str[STRSIZE -1] != 'L') tmp.push_back(n_str[STRSIZE -1]);
        if(n_str[STRSIZE -2] == 'L' && n_str[STRSIZE -1] == 'L') count++;
    }

    n_str = tmp;

    //cout << count << "\n";

    for(int i = 1; i <= K && i <= ((int)n_str.size())/2 ;i++){

        if(i == ((int)n_str.size())/2  && ((int)n_str.size())%2 == 0){
            count++;
            break;
        }
        count = count + 2;
    }

    //for(int i=0; i<(int)n_str.size(); i++){
    //    cout << n_str[i];
    //}
    //cout << "\n";

    cout << count << "\n";
}