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<iostream>
#include<string>
#include<vector>
#include <algorithm>
#include<deque>

//マクロ
#define REP(i,n) for(int i=0; i<(int)(n); ++i)
#define REPS(i,n) for(int i=1; i<=(int)(n); ++i)
#define RREP(i,n) for(int i=(int)(n)-1; i>=0; --i)
#define RREPS(i,n) for(int i=(int)(n); i>0; --i)
#define FOR(i,s,e) for(int i=(int)(s); i<(int)(e); ++i)
#define RFOR(i,s,e) for(int i=(int)(e)-1; i>=(int)(s); --i)

#define All(a) (a).begin(),(a).end()

//定数
#define INF 1e9
#define MOD (int)(1e9+7)

typedef long long ll;
typedef unsigned long long ull;
using namespace std;

//とりあえず書いておく変数
int dx[4] = { 1,0,-1,0 };
int dy[4] = { 0,1,0,-1 };

//今回の変数

//使用する関数
template<class T>bool chmax(T &a, const T &b) {
	if (a < b) { a = b; return 1; }
	return 0;
}
template<class T>bool chmin(T &a, const T &b) {
	if (a > b) { a = b; return 1; }
	return 0;
}


int main()
{
	int n,k;
	cin >> n >> k;
	char pre;
	cin >> pre;
	int ans = 0;
	int cnt = 1;
	REP(i, n-1) {
		char c;
		cin >> c;
		if (c != pre) {
			cnt++;
		}
		else {
			ans++;
		}
		pre = c;
		//cout << ans << " " << cnt << endl;
	}
	if (k <= cnt - 2) {
		ans += k * 2;
	}
	else {
		ans += (cnt - 2) * 2+1;
	}

	cout << ans << endl;
}