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<algorithm>
#include<string>
#include<cstdlib>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<math.h>
#include<stdio.h>

using namespace std;

using ll = long long;

#define rep(i, n) for (ll i = 0; i < n; i++)

const ll INF = 1LL << 60;

typedef pair<ll, double> P;

bool pairCompare(const P& firstElof, const P& secondElof)
{
	return firstElof.second < secondElof.second;
}

bool IsPrime(ll num)
{
	if (num < 2) return false;
	else if (num == 2) return true;
	else if (num % 2 == 0) return false; // 偶数はあらかじめ除く
	double sqrtNum = sqrt(num);
	for (int i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0)
		{
			// 素数ではない
			return false;
		}
	}

	// 素数である
	return true;
}

signed main() {
	ll ans = 0;

	ll n, k;
	cin >> n >> k;

	vector<ll>a(n + 1);

	rep(i, n) {
		char s;
		cin >> s;
		if (s == 'R') {
			a[i] = 1;
		}
		else {
			a[i] = 0;
		}
	}
	ll katamari = 1;
	for (ll i = 1; i < n; i++) {
		if (a[i] != a[i - 1])katamari++;
		else {
			ans++;
		}
	}
	rep(i, k) {
		if (katamari >= 3) {
			ans += 2;
			katamari -= 2;
		}
		else if (katamari == 2) {
			ans++;
			katamari--;
		}
		else break;
	}
	cout << ans;
}