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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <math.h>
#include <map>
#include <queue>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;
const int iINF = 1001001001;
const long long llINF = 1LL << 60;
int main()
{
	// ABC140-D, 1112, 52:57, WA(UNSOLVED), 
	int n, k;
	string s, invs;
	cin >> n >> k >> s;
	vector<int> v;
	int ans = 0;
	int changed = 0;
	for (int i = 1; i < n; ++i) {
		if (s[i - 1] == s[i]) ans++;
		else changed++;
	}
	changed++;
	if (changed == 1) {
		cout << ans << endl;
		return 0;
	}
	int i = 0;
	for (i = 0; i < k; ++i) {
		if (changed <= 2) break;
		ans += 2;
		changed -= 2;
	}
	if (changed == 2 && i != k) ans++;
	cout << ans << endl;
	return 0;
}