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>
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define RFOR(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define rep(i,n) FOR(i,0,(n))
#define REPS(i,n) FOR(i,1,(n)+1)
#define RREP(i,n) RFOR(i,0,(n))
#define rng(x) (x).begin(),(x).end()
#define rrng(x) (x).rbegin(),(x).rend()
#define debug(x) cout << #x << " = " << (x) << endl;
#define SORT(x) sort(rng(x))
#define RSORT(x) sort(rrng(x))
#define SUM(x) accumulate(rng(x),0)
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define uni(x) x.erase(unique(rng(x)),x.end())
#define SP << " " <<
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using pii = pair<int, int>;
using tiii = tuple<int, int, int>;
const int mod = 1000000007;
const double EPS = 1e-9;
const int INF = (1<<30)-1;
const ll INFLL = (1LL<<62)-1;
const int dx[4] = {1,0,-1,0};
const int dy[4] = {0,1,0,-1};
//cout << fixed << setprecision(10);

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}

int main () {
  int n, k;
  string s;
  cin >> n >> k >> s;

  int cnt = 0;
  rep(i,n-1) {
    if (s[i] == s[i+1]) cnt++; 
  }

  cout << min(n-1, cnt+2*k) << endl;

  return 0;
}