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: Wrong Answer)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n)  for(long long i=0;i<(long long)(n);i++)
#define Rep(i,k,n) for(long long i=k;i<(long long)(n);i++)
#define all(a) a.begin(),a.end()
typedef long long ll;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> PP;
typedef priority_queue<ll> PQ;
typedef priority_queue<ll,vector<ll>,greater<ll>> SMALLPQ;
using vi=vector<ll>;
using vvi=vector<vector<ll>>;
using vc=vector<char>;
using vvc=vector<vector<char>>;
const int inf=1000000007;
const int mod=1000000007;
void chmin(auto &a,auto b){if(b<a)a=b;}
void chmax(auto &a,auto b){if(a<b)a=b;}
int main(){
  int n,k,ans=0;cin>>n>>k;
  string s;cin>>s;
  if(s[0]=='L')ans++;
  if(s[n-1]=='R')ans++;
  rep(i,n-1)if(s[i]=='R'&&s[i+1]=='L')ans++;
  if(ans-2*k<=1)cout<<n-1<<endl;
  else cout<<n-(ans-2*k)<<endl;
}