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<iostream>
#include<sstream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<iomanip>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<string>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<map>
#include<set>
#define mem(a,b) memset(a,b,sizeof(a))
#define random(a,b) (rand()%(b-a+1)+a)
#define ll long long
#define ull unsigned long long
#define e 2.71828182
#define Pi acos(-1.0)
#define ls(rt) (rt<<1)
#define rs(rt) (rt<<1|1)
#define lowbit(x) (x&(-x))
using namespace std;
const int MAXN=101;
int A[MAXN],B[MAXN],C[MAXN];
int read()
{
	int s=1,x=0;
	char ch=getchar();
	while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();}
	while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();}
	return x*s;
}
struct node
{
	int ma;
	int x,y;
	node(int m,int xx,int yy)
	{
		ma=m;x=xx;y=yy;
	}
	bool operator < (const node& a) const
	{
		return ma<a.ma;
	}
};
string str;
int main()
{
	int n=read(),k=read();
	cin>>str;
	int ans=0;
	for(int i=1;i<n;++i) if(str[i]==str[i-1]) ans++;
	ans=min(ans+2*k,n-1);
	cout<<ans<<endl;
}