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: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;
#define REP(i,a) for(int i = 0; i < (a);i++)
#define ALL(a) (a).begin(),(a).end()
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const int MOD = 1e9 + 7;
using Graph = vector < vector<int>>;
long int sum[200020] = {};
/*第二引数で第一引数を割ったときの切り上げの計算*/
long long int maxtime(long long int x,long long int y) {
	return(x + y-1) / y;

}
/*最大公約数*/
long long int lcm(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;

	if (number2 > number1) {
		m = number2;
		n = number1;
	}
	long long int s = -1;
	while (s != 0) {
		s = m % n;
		m = n;
		n = s;
	}
	return m;
}
/*最大公倍数*/
long long int gcd(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;
	return m / lcm(m, n) * n;
}

int main() {
	int n, k;
	cin >> n >> k;
	string N;
	cin >> N;
	vector<int> L;
	vector<int> R;
	int sum = 0;
	char C = ' ';
	int l = 0;
	int r = 0;
	for (int i = 0; i < n; i++) {
		if (i == 0) {
			C = N[i];
			sum = sum + 1;
			continue;
		}

		if (N[i] == C) {
			sum = sum + 1;
			if (i == n - 1) {
				if (C == 'L') {
					L.push_back(sum);
					l = l + 1;
				}
				else {
					R.push_back(sum);
					r = r + 1;
				}
			}
		}
		else {
			if (C == 'L') {
				L.push_back(sum);
				C = 'R';
				sum = 1;
				l = l + 1;
			}
			else {
				R.push_back(sum);
				C = 'L';
				sum = 1;
				r = r + 1;
			}
		}
	}
	sum = 0;
	for (int i = 0; i < r; i++) {
			sum = sum + R[i] - 1;
	}
	for (int i = 0; i < l; i++) {
			sum = sum + L[i] - 1;
	}
	if (min(l, r) <= k) {
		sum = N.size() - 1;
	}
	else {
		sum = sum + k + k;
	}
	cout << sum;
}