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;
 
typedef long long ll;
typedef pair<ll,ll> P;
 
long long int INF = 3e18;
double Pi = 3.1415926535897932384626;
 
vector<ll> G[550010];
//vector<P> tree[500010];
priority_queue <ll> pql;
priority_queue <P> pqp;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop
 
int dx[8]={1,0,-1,0,1,1,-1,-1};
int dy[8]={0,1,0,-1,1,-1,-1,1};
char dir[] = "DRUL";
//ll bit[500005];
//↓,→,↑,←
 
 
#define p(x) cout<<x<<"\n";
#define el cout<<endl;
#define pe(x) cout<<(x)<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define CLEAR(a) a = decltype(a)(); 
 
 
ll mod = 1000000007;
/*
2000000011 is prime
2000000033 is prime
*/

ll rui(ll number1,ll number2){
 
    if(number2 == 0){
        return 1;
    }else{
        ll number3 = rui(number1,number2 / 2);
        number3 *= number3;
        number3 %= mod;
        if(number2%2==1){
            number3 *= number1;
            number3 %= mod;
        }
        return number3;
    }
}
ll gcd(ll number1,ll number2){
 
    if(number1 > number2){
        swap(number1,number2);
    }
    if(number1 == 0 || number1 == number2){
        return number2;
    }else{
        return gcd(number2 % number1,number1);
    }
}
void YES(bool condition){
 
    if(condition){
        p("YES");
    }else{
        p("NO");
    }
    return;
}
void Yes(bool condition){
 
    if(condition){
        p("Yes");
    }else{
        p("No");
    }
    return;
}

/*
ll fact[800005],rfact[800005];

void c3_init(){
    fact[0] = rfact[0] = 1;
    for(ll i=1; i<=800000; i++){
        fact[i] = (fact[i-1]*i) % mod;
    }
    rfact[800000] = rui(fact[800000],mod - 2);
    for(ll i=800000; i>=1; i--){
       rfact[i-1] = rfact[i] * i;
       rfact[i-1] %= mod;
    }
    return;}
ll c3(ll n,ll r){
    return (((fact[n] * rfact[r]) % mod) * rfact[n-r]) % mod;}
*/
 
ll n,m,num,sum,a,b,c,d,e,g,h,w,i,j,q,r,l;
ll k,ans;
ll x[500005],y[500005],z[500005];
ll res;
char s[500005];

int main(){
    cin >> n >> k;
    cin >> s;
    for(int i=0;i<n-1;i++){
        if(s[i] == s[i+1]){
            res++;
        }else{
            ans += res;
            res = 0;
        }
    }
    ans += res;
    res = 0;
    ans += 2 * k;
    ans = min(ans, n - 1);
    p(ans);
    return 0;
}