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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vvi = vector<vi>;
using vvl = vector<vl>;
const ll INF = 1LL << 60;
const ll MOD = 1000000007;
template <class T>
bool chmax(T &a, const T &b) {
    return (a < b) ? (a = b, 1) : 0;
}
template <class T>
bool chmin(T &a, const T &b) {
    return (b < a) ? (a = b, 1) : 0;
}
template <class C>
void print(const C &c, std::ostream &os = std::cout) {
    std::copy(std::begin(c), std::end(c), std::ostream_iterator<typename C::value_type>(os, " "));
    os << std::endl;
}

int main() {
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    int ret = 0;
    char curr = s[0];
    for (int i = 1; i < n; ++i) {
        if (k == 0)
            break;
        if (s[i] == s[i - 1]) {
            // ret++;
        } else {
            k--;
            for (int j = i; j < n; ++j) {
                if (s[j] != s[i - 1])
                    s[j] = s[i - 1];
                // ret++;
                else {
                    i = j;
                    break;
                }
            }
        }
    }
    for (int i = 1; i < n; ++i) {
        if (s[i] == s[i - 1])
            ret++;
    }
    cout << ret << "\n";
    return 0;
}