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 <bits/stdc++.h>
using namespace std;
#define REP(i,n) for(int i=0;i<(n);i++)
#define N 500000
#define INF 1000000000
typedef long long ll;
#define ALL(v) (v).begin(),(v).end()
#define SZ(x) int(x.size())
#define IN(a) cin>>(a)
#define OUT(a) cout<<(a)<<endl
typedef pair<int,int> P;
const int MAX = 1000000;
const int MOD = 1000000007;
int main(){
	int n,k;
    cin>>n>>k;
    string s;
    cin>>s;
    int ans=0;
    REP(i,n-1){
        if(s[i]==s[i+1])ans++;
    }
    int tmp=n-ans;
    if(tmp/2>k){
        ans+=2*k;
    }else{
        ans=n-1;
    }
    OUT(ans);
	return 0;
}