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

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=(s);i<=(n);i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
#define REPR(i,s,n) for(int i=(s);i>=(n);i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define Eunique(v) v.erase(unique(all(v)),v.end())
#define Eback(s) s.erase(s.end()-1,s.end())
#define rev(v) reverse(all(v))
#define minvec(v) *min_element(all(v))
#define maxvec(v) *max_element(all(v))
#define sumvec(v) accumulate(all(v),0LL)
#define mapmin(v) v.rbegin()->first
#define mapmax(v) v.begin()->first
#define pb push_back
#define pf push_front
#define m_p make_pair
#define DOUBLE fixed << setprecision(15)
#define OK cerr<<"OK\n"
#define OK1 cerr<<"OK1\n"
#define OK2 cerr<<"OK2\n"
#define sz(s) (long long)s.size()
#define INF ((1LL<<62)-(1LL<<31))
#define zero(x,n) setw(x) << setfill('0') << n
#define endl '\n'

typedef long long ll;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<long long> vll;
typedef vector<vll> vvll;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<string> vs;

typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
typedef vector<pair<int,int>> vpii;
typedef vector<pair<ll,ll>> vpll;

const double pi = acos(-1.0);
const ll mod=1000000007;
const ll mod2=998244353;

template<class A, class B>
ostream& operator<<(ostream& ost, const pair<A, B>&p) {
	ost << "{" << p.first << ", " << p.second << "} ";
	return ost;
}

template<class T>
ostream& operator<<(ostream& ost, const vector<T>&v) {
	ost << "{";
	for (int i = 0; i<(int)v.size(); i++) {
		if (i)ost << " ";
		ost << v[i];
	}
	ost << "} \n";
	return ost;
}

template<class A, class B>
ostream& operator<<(ostream& ost, const map<A, B>&v) {
	ost << "{";
	for (auto p:v) {
		ost << "{" << p.first << ", " << p.second << "} ";
	}
	ost << "} ";
	return ost;
}

void out(){cout << "\n";}
template<class T,class... Args>
void out(const T &x,const Args &... args){
	cout << x << " ";
	out(args...);
}

template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
	for(T& e: vec) is >> e;
	return is;
}

template<typename T>
string join(vector<T>& vec, string sep=" "){
	stringstream ss;
	for(int i=0;i<(int)vec.size();i++){
		ss << vec[i] << (i+1==(int)vec.size()?"\n":sep);
	}
	return ss.str();
}

template<class T>
inline bool chmax(T& a, T b){if(a<b){a=b;return true;} return false;}

template<class T>
inline bool chmin(T& a, T b){if(a>b){a=b;return true;} return false;}

void YES(bool b) {cout << ((b)?"YES\n":"NO\n");}
void Yes(bool b) {cout << ((b)?"Yes\n":"No\n");}
void yes(bool b) {cout << ((b)?"yes\n":"no\n");}
void Yay(bool b) {cout << ((b)?"Yay!\n":":(\n");}

ll powmod(ll a,ll b){ll c=1; while(b>0){ if(b&1){ c=a*c%mod; } a=a*a%mod; b>>=1; } return c;}
ll nCrmod(ll n,ll r){ll x=1,y=1; for(ll i=0;i<r;i++) {x=x*(n-i)%mod; y=y*(i+1)%mod; } return x*powmod(y,mod-2)%mod;}
ll gcd(ll x, ll y) {return __gcd(x,y);}
ll lcm(ll x, ll y) {return x/__gcd(x,y)*y;}

int main()
{
	ll n,k;
	string s;
	cin >> n >> k >> s;
	ll score=0;
	rep(i,n-1){
		if(s[i]==s[i+1]) score++;
	}
	ll ans=min(score+2*k,n-1);
	cout << ans << endl;
	return 0;
}