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, n) for (int i = 0; i < (int)(n); i++)
#define REP2(i, a, b) for (int i = (a); i < (int)(b); i++)
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef long long ll;
#define ALL(V) V.begin(), V.end()


int Main(int N, int K, string S)
{
    int ans = 0;
    VI chunks(1, -1);
    int counter = 0;
    char before = S[0];
    chunks[0]++;
    REP2(i, 1, N) {
        if(S[i] == before) {
            chunks[counter]++;
        }
        else {
            counter++;
            chunks.push_back(0);
        }
        before = S[i];
    }
    int subsum = accumulate(ALL(chunks), 0);
    if(counter == 1) {
        return subsum + 1;
    }
    else if (counter == 0) {
        return subsum;
    }
    else {
        return subsum + 2 * min(K, counter/2);
    }
}

int main(void){
    int N, K;
    cin >> N >> K;
    string S;
    cin >> S;
    cout << Main(N, K, S) << endl;
    return 0;
}