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>
#define ff first
#define ss second
#define len(x) (int)(x.size())
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using ii = pair<int, int>;

int main(){
	ios::sync_with_stdio(0);
	int n, k;
	cin >> n >> k;
	string st;
	cin >> st;
	int i = 0;
	vector<int> vs;
	while(i<n){
		int c = st[i];
		int w = 0;
		while(i<n and st[i]==c) ++w, ++i;
		vs.push_back(w);
	}
	while(k--){
		if(len(vs)<=2){
			cout << n-1 << endl;
			return 0;		
		}
		int w = vs.back(); vs.pop_back();
		w += vs.back(); vs.pop_back();
		w += vs.back(); vs.pop_back();
		vs.push_back(w);
	}
	int ans = 0;
	for(auto x:vs) ans += x-1;
	cout << ans << endl;
}