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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <cstdio>
#include <cmath>
#include <tuple>
#include <iomanip>
#include <numeric>
#include <unordered_map>
#include <sstream>
#include<limits.h>
#include<float.h>
#include<list>
#include <array>
#include <complex>
#include<stdio.h>
#include<string.h>
#include <bitset>
#include<assert.h>
#include<random>
using namespace std;
#define int long long

#define I32_MAX 2147483647
#define I64_MAX 9223372036854775807LL
#define I64_MAX2 1223372036854775807LL
#define INF I64_MAX2
#define MOD 1000000007
// #define MOD 998244353 
#define MEM_SIZE 10000
#define DEBUG_OUT true
#define ALL(x) (x).begin(), (x).end()

template<typename T> void DEBUG(T e){if(DEBUG_OUT == false)return; std::cout << e <<" ";}
template<typename T> void DEBUG(const std::vector<T>& v){if(DEBUG_OUT == false)return;for(const auto& e : v){std::cout<< e << " "; } std::cout << std::endl;}
template<typename T> void DEBUG(const std::vector<std::vector<T> >& vv){if(DEBUG_OUT == false)return;for(const auto& v : vv){ DEBUG(v); } }
template<class T,class... Ts> void DEBUG(T d, Ts... e){if(DEBUG_OUT == false)return;DEBUG(d);DEBUG(e...);}
template <class T> void corner(bool flg, T hoge) {if (flg) {cout << hoge << endl; abort();}}
template< typename T1, typename T2 > inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template< typename T1, typename T2 > inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

void solve(void)
{
  int N,K;cin>>N>>K;
  string str;cin>>str;
  int cnt_dc = 0;
  for (int i = 0; i < N-1; i++)
  {
    if(str[i+1] != str[i])cnt_dc++;
  }
  cout<<min(N - cnt_dc -1 + K*2,N-1)<<endl;
  
  return;
}
int32_t main(int32_t argc, const char *argv[])
{
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);

  std::cout << std::fixed;
  std::cout << std::setprecision(11);
  solve();

  return 0;
}