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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
// #include <atcoder/all>
#define ll long long int
#define ld long double
#define yorn(f) cout<<((f)?"Yes":"No")<<endl;
#define YORN(f) cout<<((f)?"YES":"NO")<<endl;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repi(i, n) for (ll i = 1; i < (ll)(n); i++)
#define all(x) (x).begin(), (x).end()
#define put(x) cout << x << endl;
#define println(x) cout << x << endl;
using namespace std;
// using namespace atcoder;
void Main()
{
    ll n, K;
    cin >> n >> K;

    string S;
    cin >> S;

    ll res = 0;

    ll baseHappy = 0;
    rep(i, n) {
        if (S[i] == 'L') {
            if (0 <= i -1 &&S[i - 1] == 'L') {
                baseHappy++;
            }
        } 
        if (S[i] == 'R') {
            if (i + 1 < n && S[i + 1] == 'R') {
                baseHappy++;
            }
        } 
    }

    for (char startChar: { 'L', 'R' }) {
        char nonStartChar = startChar == 'L' ? 'R' : 'L';
        char ch[2] = { startChar, nonStartChar };
        ll l = 0, r = n - 2;

        ll nHappy = baseHappy;
        ll op = 0;
        while (l < r) {
            while (l + 1 < S.size() && (S[l] != ch[0] || S[l + 1] != ch[1])) {
                l++;
            }
            while (0 <= r && (S[r] != ch[1] || S[r + 1] != ch[0])) {
                r--;
            }

            if (l < r) {
                nHappy += 2;
                op++;
            }

            if (op >= K) {
                break;
            }

            swap(ch[0], ch[1]);
        }

        if (op < K && S[0] == nonStartChar) {
            nHappy++;
            op++;
        }

        if (op < K && S[n - 1] == nonStartChar) {
            nHappy++;
            op++;
        }

        res = max(res, nHappy);
    }

    put(res);
}
signed main(){ Main();return 0;}