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;

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

template<typename T>
ostream& operator << (ostream& ostr, vector<T>& vec){
    if(!vec.empty()){
        ostr << vec.front();
        for(auto itr = ++vec.begin(); itr != vec.end(); itr++)
            ostr << " " << *itr;
    }
    return ostr;
}

template<typename T>
ostream& operator << (ostream& ostr, vector<vector<T>>& mat){
    if(!mat.empty()){
        ostr << mat.front();
        for(auto itr = ++mat.begin(); itr != mat.end(); itr++)
            ostr << endl << *itr;
    }
    return ostr;
}

template<typename T, typename U>
ostream& operator << (ostream& ostr, pair<T,U>& p){
    ostr << p.first << " " << p.second;
    return ostr;
}

template<typename T, typename U>
ostream& operator << (ostream& ostr, vector<pair<T,U>>& mat){
    if(!mat.empty()){
        ostr << mat.front();
        for(auto itr = ++mat.begin(); itr != mat.end(); itr++)
            ostr << endl << *itr;
    }
    return ostr;
}

typedef long long ll;
typedef vector<ll> vll;
// vll x(n); x[i]=a;
typedef vector<pair<ll,ll>> vpll;
// vpll x(n); x[i]={a,b}; x[i].first/second;
typedef vector<vector<ll>> vvll;
// vvll x(m,vi(n)); x[i][j]=a;

#define REP(i,n) for(ll i=0; i<(n); i++)
#define RREP(i,n) for(ll i=(n)-1; i>=0; i--)
#define FOR(i,a,b) for(ll i=(a); i<(b); i++)
#define FORR(i,a,b) for(ll i=(b)-1; i>=(a); i--)
#define dump(x) cout << #x << " = " << (x) << endl
#define all(x) (x).begin(),(x).end()
// sort(all(x));

int main(){
    ll n, k, ans = 0, cnt1 = 0, cnt2 = 0;
    string s;
    cin >> n >> k;
    cin >> s;

    if(s[0] == 'L') cnt1 += 1;
    REP(i,n-1) if(s[i] == 'R' && s[i+1] == 'L') cnt2 += 1;
    if(s[n-1] == 'R') cnt1 += 1;
    ans = n-cnt1-cnt2*2;

    while(k > 0 && cnt2 > 0){
        ans += 2;
        k--;
        cnt2--;
    }
    if(cnt2 == 0){
        if(cnt1 == 0) ans--;
        else if(cnt1 == 2 && k > 0) ans++;
    }
    cout << ans << endl;
    return 0;
}