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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#include <iostream>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <math.h>
#include <sstream>
#include <numeric>
#include <cctype>
#include <bitset>
#include <cassert>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define SIZE 100005
#define INF 1000000000000000LL
using pint = pair<int,int>;
using vec = vector<int>;
using ll=long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }


int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n,k;cin>>n>>k;
    string s; cin>>s;

    int cnt=0;
    for(int i=0; i<n-1; i++){
        if(s[i]==s[i+1]) cnt++;
    }
    int ans = min(cnt+2*k,n-1);
    cout<<ans<<endl;
    }