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;
#define repi(n) for (int i = 0; i < (n); i++)
#define repj(n) for (int j = 0; j < (n); j++)
#define repk(n) for (int k = 0; k < (n); k++)
#define repl(n) for (int l = 0; l < (n); l++)
#define rep(i, n) for (int i = 0; (i) < (n); i++)
#define repr(i, a, b) for (auto i = (a); i < (b); i++)
#define repv(itr) for (auto&& v : (itr))
#define updatemax(t, v) (t = max((t), (v)))
#define updatemin(t, v) (t = min((t), (v)))
const int dx[] = {-1, 0, 0, 1, -1, -1, 1, 1};
const int dy[] = {0, -1, 1, 0, -1, 1, -1, 1};
template <typename T>
T minptr(T begin, T end) {
  T re = begin;
  for (T i = begin + 1; i != end; i++) {
    if (*i < *re) re = i;
  }
  return re;
}
template <typename T>
T maxptr(T begin, T end) {
  T re = begin;
  for (T i = begin + 1; i != end; i++) {
    if (*i > *re) re = i;
  }
  return re;
}
int __vmax(int x) {
  return INT_MAX;
}
double __vmax(double x) {
  return 1e+300;
}
ll __vmax(ll x) {
  return LLONG_MAX;
}
int __vmin(int x) {
  return INT_MIN;
}
double __vmin(double x) {
  return -1e+300;
}
ll __vmin(ll x) {
  return LLONG_MIN;
}
ll modpow(ll a, ll b, ll m) {
  ll re = 1, k = 1;
  while (k <= b) {
    if (b & k) {
      re *= a;
      re %= m;
    }
    k = k << 1;
    a *= a;
    a %= m;
  }
  return re;
}
ll modinv(ll a, ll m) {
  ll b = m, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0) u += m;
  return u;
}
template <typename T>
vector<T> lis(T begin, T end, bool allowequal = false, bool lds = false) {
  using V = typename iterator_traits<T>::value_type;
  int n = end - begin;
  vector<V> a(n, lds ? __vmin(*begin) : __vmax(*begin));
  vector<int> id(n);
  if (lds && allowequal) {
    for (int i = 0; i < n; i++) {
      id[i] = n - 1 -
              (lower_bound(a.rbegin(), a.rend(), begin[i]) - 1 - a.rbegin());
      a[id[i]] = begin[i];
    }
  } else if (lds) {
    for (int i = 0; i < n; i++) {
      id[i] = n - 1 -
              (upper_bound(a.rbegin(), a.rend(), begin[i]) - 1 - a.rbegin());
      a[id[i]] = begin[i];
    }
  } else if (allowequal) {
    for (int i = 0; i < n; i++) {
      id[i] = upper_bound(a.begin(), a.end(), begin[i]) - a.begin();
      a[id[i]] = begin[i];
    }
  } else {
    for (int i = 0; i < n; i++) {
      id[i] = lower_bound(a.begin(), a.end(), begin[i]) - a.begin();
      a[id[i]] = begin[i];
    }
  }
  int m = *maxptr(id.begin(), id.end());
  vector<T> re(m + 1);
  for (int i = n - 1; i >= 0; i--) {
    if (id[i] == m) re[m--] = begin + i;
  }
  return re;
}

int main() {
  int n, k;
  string s;
  cin >> n >> k >> s;
  vector<int> ls, rs;
  char pre = s[0];
  int prelen = 0;
  repi(n) {
    if (s[i] == pre) {
      prelen++;
    } else {
      (pre == 'L' ? ls : rs).push_back(prelen);
      pre = s[i];
      prelen = 1;
    }
  }
  (pre == 'L' ? ls : rs).push_back(prelen);
  int ll = ls.size(), rl = rs.size();
  if (s[0] == s[n - 1]) {  // XYX...YX
    cout << n - 1 - max((s[0] == 'L' ? rl : ll) - k, 0) * 2 << endl;
  } else {  // XYXY...X
    cout << n - max((ll - k) * 2, 1) << endl;
  }
}