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<set>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<map>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;

typedef long long ll;
int mod=1e9+7;
const int maxn=1e5+10;
char  a[maxn];
priority_queue<int>q;
vector<int>v;
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    scanf("%s",a+1);
    int cnt1=0;
    int cnt2=0;

    for(int i=1;i<=n;i++)
    {
        if(a[i]=='L'&&a[i-1]!='L')
        {
            cnt1++;
        }
        else if(a[i]=='R'&&a[i-1]!='R')
        {
            cnt2++;
        }
    }
    if(k>=min(cnt1,cnt2))
    {
        printf("%d\n",n-1);
    }
    else
    {
        printf("%d\n",min(2*k+n-cnt1-cnt2,n-1));
    }
    return 0;
}