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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

#define REP(i, a, b) for (int i = a; i < b; i++)
#define rep(i, n) REP(i, 0, n)
#define all(x) (x).begin(), (x).end()

typedef long long ll;
typedef unsigned long long ull;

struct prs {
    int x1, x2, sum;

    // 演算子オーバーロードで比較関数を定義
    bool operator<(const prs& another) const {
        if (sum == another.sum) {
            return x1 > another.x1;
        } else {
            return sum > another.sum;
        }
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N, K;
    cin >> N >> K;
    string S;
    cin >> S;
    int score = 0;
    rep(i, N - 1) {
        if (S[i] == S[i + 1]) {
            score++;
        }
    }
    vector<prs> cting;
    int x1 = 0, x2 = 0;
    rep(i, N - 1) {
        if (S[i] == S[i + 1]) {
            x2 = i + 1;
        } else {
            prs p = {.x1 = x1, .x2 = x2, .sum = x2 - x1 + 1};
            cting.push_back(p);
            x1 = i + 1;
            x2 = i + 1;
        }
    }
    sort(all(cting));
    for (int i = 0; i < K; i++) {
        if (cting[i].x1 == 0) {
            continue;
        } else if (cting[i].x2 == N - 1) {
            score += 1;
            continue;
        } else {
            score += 2;
        }
    }
    cout << score << endl;
    // for(auto c :cting){
    //     printf("%d %d %d\n", c.x1, c.x2, c.sum);
    // }
    return 0;
}