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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    int n,k,a,b,ans;
    ans = 0;
    cin>>n>>k;
    string s;cin>>s;
    char m;
    for(int o=0;o<k;o++){
        m=s[0];
        //cout<<s[0]<<endl;
        for(int i=0;i<n;i++){
            
            if(i==n-2&&s.at(i)==m){
                cout<<n-1<<endl;
                return 0;
            }
            
            if(m!=s.at(i)){
                a = i;
                cout<<i<<endl;
                break;
            }
        }
        
                for(int j=n-2;j>=0;j--){
                    if(m!=s.at(j))b=j;
                    break;
                }
                
                //cout<<a<<" "<<b<<endl;
                for(int i=0;i<(a+b)/2;i++){
                    char v = s.at(i);
                    s.at(i+a)=s.at(b-i);
                    s.at(b-i)=v;
                }
                
                //cout<<s<<endl;
                for(int i=a;i<=b;i++){
                    if(s.at(i)=='R')s.at(i)='L';
                    else s.at(i)='R';
                }
    }
    char x = s.at(0);
    for(int i=1;i<n;i++){
        if(x==s.at(i))ans++;
        else{
            x = s.at(i);
        }
    }
    cout<<ans<<endl;
}