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>

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define LL long long
#define pii pair<int,int>
#define SZ(x)  (int)x.size()
#define all(x) x.begin(),x.end()

using namespace std;

LL gcd(LL a, LL b) {return b ? gcd(b, a % b) : a;}
LL lcm(LL a, LL b) {return a / gcd(a, b) * b;}
LL powmod(LL a, LL b, LL MOD) {LL ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
const int N = 5e5 + 3;
const LL mod = 1e9 + 7;
int n,k;
char a[N];
int cmp(pii a,pii b){
    return a.se-a.fi>b.se-b.fi;
}
int main() {
	ios::sync_with_stdio(false);
    cin>>n>>k>>a+1;
    int pre,last,x,y;
    pre=last=0;
    x=y=0;
    vector<pii>p,q;
    for(int i=1;i<=n;i++){
        if(a[i]=='L'&&pre){
            p.pb({pre,last});
            pre=0;
            last=0;
        }else if(a[i]=='R'){
            if(pre){
                last=i;
            }else{
                pre=last=i;
            }
        }

        if(a[i]=='R'&&x){
            q.pb({x,y});
            x=y=0;
        }else if(a[i]=='L'){
            if(!x){
                x=y=i;
            }else y=i;
        }

    }
    if(pre){
        p.pb({pre,last});
    }
    if(x){
        q.pb({x,y});
    }
    sort(all(p),cmp);
    sort(all(q),cmp);
    if(SZ(p)<SZ(q))
        for(int i=0;i<min(k,SZ(p));i++){
            for(int j=p[i].fi;j<=p[i].se;j++){
                a[j]='L';
            }
        }
    else{
        for(int i=0;i<min(k,SZ(q));i++){
           // cout<<q[i].fi<<' '<<q[i].se<<endl;
            for(int j=q[i].fi;j<=q[i].se;j++){
                a[j]='R';
            }
        }
    }
    int ans=0;
    for(int i=1;i<=n;i++){
        if(a[i]=='L'&&i>1){
            if(a[i-1]=='L')ans++;
        }else if(a[i]=='R'&&i<n){
            if(a[i+1]=='R')ans++;
        }
    }
    cout<<ans<<endl;
	return 0;
}