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)

#include<bits/stdc++.h>
#include<algorithm>
using namespace std;

#define ll long long
#define ld long double

#define rep(i,a,b) for(ll i=a;i<b;i++)
#define repb(i,a,b) for(ll i=a;i>=b;i--)
#define repA(i,A)   for(auto i:A)

#define err() cout<<"=================================="<<endl;
#define errA(A) for(auto i:A)   cout<<i<<" ";cout<<endl;
#define err1(a) cout<<#a<<" "<<a<<endl
#define err2(a,b) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<endl
#define err3(a,b,c) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<" "<<#c<<" "<<c<<endl
#define err4(a,b,c,d) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<" "<<#c<<" "<<c<<" "<<#d<<" "<<d<<endl

#define mp make_pair
#define pb push_back
#define all(A)  A.begin(),A.end()
#define allr(A)	A.rbegin(),A.rend()
#define ft first
#define sd second

#define pll pair<ll,ll>
#define V vector<ll>
#define S set<ll>
#define VV vector<V> 
#define Vpll vector<pll>

#define endl "\n"

const ll logN = 20;
const ll M = 1000000007;
const ll INF = 1e12;
#define PI 3.14159265

const ll N = 100005;

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	ll n,k;
	string s;
	cin>>n>>k>>s;
	ll a=0,b=0;
	s = 'L' + s + 'R';
	rep(i,0,n+1){
		if(s[i]=='L' and s[i+1]=='R')	a++;
	}
	rep(i,1,n+1){
		if(s[i]=='R' and s[i-1]=='R')	b++;
		if(s[i]=='L' and s[i+1]=='L')	b++;
	}
	rep(i,0,k){
		if(a==1)	break;
		else if(a==2){
			a--;
			b++;
		}
		else{
			a--;
			b+=2;
		}
	}
	cout<<b<<endl;
	return 0;
}