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 <bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define FOR(i,s,n) for(int i = s; i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(n) (n).begin(), (n).end()
#define RALL(n) (n).rbegin(), (n).rend()
#define ATYN(n) cout << ( (n) ? "Yes":"No") << endl;
#define CFYN(n) cout << ( (n) ? "YES":"NO") << endl;
using ll = long long;
using ull = unsigned long long;

int main(void)
{
    IOS
    int n, k;
    cin >> n >> k;
    string s; cin >> s;
    int ans = 0;


    REP(i,n) {
        if (i != n-1 && s[i] == 'r') {
            if (s[i] == s[i+1]) ans++;
        } else if (i != 0) {
            if (s[i] == s[i-1]) ans++;
        }
    }

    ans = min(ans + k * 2,n-1);

    cout << ans << endl;

    return 0;
}