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<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<functional>
#include<map>
#include<cstdlib>
#include<complex>
#include<numeric>
#include<set>
#include<deque>

using namespace std;
using Int = long long;
Int INF = 1LL << 60;
const Int MOD = 1000000000 + 7;

template<int Modulo = MOD> struct Mint {
	Int val;
	constexpr Mint(Int v = 0) noexcept : val(v% Modulo) {
		if (val < 0) v += Modulo;
	}
	constexpr int getmod() { return Modulo; }
	constexpr Mint operator - () const noexcept { return val ? Modulo - val : 0; }
	constexpr Mint operator + (const Mint& r) const noexcept { return Mint(*this) += r; }
	constexpr Mint operator - (const Mint& r) const noexcept { return Mint(*this) -= r; }
	constexpr Mint operator * (const Mint& r) const noexcept { return Mint(*this) *= r; }
	constexpr Mint operator / (const Mint& r) const noexcept { return Mint(*this) /= r; }
	constexpr Mint& operator += (const Mint& r) noexcept {
		val += r.val;
		if (val >= Modulo) val -= Modulo;
		return *this;
	}
	constexpr Mint& operator -= (const Mint& r) noexcept {
		val -= r.val;
		if (val < 0) val += Modulo;
		return *this;
	}
	constexpr Mint& operator *= (const Mint& r) noexcept {
		val = val * r.val % Modulo;
		return *this;
	}
	constexpr Mint& operator /= (const Mint& r) noexcept {
		Int a = r.val, b = Modulo, u = 1, v = 0;
		while (b) {
			Int t = a / b;
			a -= t * b; swap(a, b);
			u -= t * v; swap(u, v);
		}
		val = val * u % Modulo;
		if (val < 0) val += Modulo;
		return *this;
	}
	constexpr bool operator == (const Mint& r) const noexcept {
		return this->val == r.val;
	}
	constexpr bool operator != (const Mint& r) const noexcept {
		return this->val != r.val;
	}
	friend ostream& operator << (ostream& os, const Mint<Modulo>& x) noexcept {
		return os << x.val;
	}
	friend istream& operator >> (istream& is, Mint<Modulo>& x) noexcept {
		Int tmp; is >> tmp;
		x = Mint<Modulo>(tmp);
		return is;
	}
	friend constexpr Mint<Modulo> modpow(const Mint<Modulo>& a, Int n) noexcept {
		Mint res(1), tmp = a;
		while (n > 0) {
			if (n & 1) res *= tmp;
			tmp *= tmp;
			n >>= 1;
		}
		return res;
	}
};

using mint = Mint<>;

int main() {

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	Int N, K;
	cin >> N >> K;

	string S;
	cin >> S;

	Int sector = 0;
	char temp = S[0];

	for (Int i = 1; i < N; i++) {
		if (S[i] == temp)continue;
		else {
			sector++;
			temp = S[i];
		}
	}

	Int ans;
	ans = (N - 1) - max(0LL, sector - 2 * K);

	cout << ans << "\n";
}