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>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define all(a) a.begin(),a.end()
#define P pair<int,int>
#define F first
#define S second
#define mod 1000000007
#define INF 9000000000000000000
signed main(void){
    int n,k;cin>>n>>k;
    string s;cin>>s;
    int ans=0;
    s+='y';
    rep(i,n-1){
        if(s[i]=='R'&&s[i+1]=='R')ans++;
    }
    rep(i,n-1){
        if(s[n-i-1]=='L'&&s[n-i-2]=='L')ans++;
    }
    string x="";
    rep(i,n){
        if(s[i]==s[i+1])continue;
        x+=s[i];
    }
    //cout<<x<<endl;
    int c=0;
    rep(i,n-2){
        if(x[i+1]=='L')c++;
    }
    if(n==1)ans=0;
    if(n==2)ans=1;
    else{
    if(c>=k)ans+=2*k;
    else{
        if(k-c>=2)ans=n-1;
        else{
            if(s[0]=='L'&&s[n-1]=='L'&&x[1]=='R')ans=n-2;
            else ans=n-1;
        }
    }
    }
    cout<<ans<<endl;
}