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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define mod 1000000007
#define MAX 1000000000000000
#define all(v) v.begin(),v.end()
#define rep(i,a,b) for(i=(ll)a;i<(ll)b;i++)
#define revrep(i,a,b) for(i=(ll)a;i>=(ll)b;i--)
#define ii pair<ll,ll>
#define MP make_pair
#define pb push_back
#define f first
#define se second
#define ll long long int
#define vi vector<ii>
ll modexp(ll a,ll b){ ll res = 1; while(b > 0){  if(b & 1) res = (res * a)%mod;  a = (a * a)%mod;  b/=2;  }  return res; }
#define rs resize
typedef tree< ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update > OST;
#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <typename Arg1>
    void __f(const char* name, Arg1&& arg1){
        cout << name << " : " << arg1 << endl;
    }
    template <typename Arg1, typename... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args){
        const char* comma = strchr(names + 1, ','); cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
    }
#else
#define trace(...)
#endif

const ll N = 1e5 + 2;
ll i,n,k,j,ans;
string s;
int main()
{
    std::ios_base::sync_with_stdio(false); cin.tie(NULL);
    cin>>n>>k>>s;
    rep(i,0,n){
        j = i + 1;
        while(j < n && s[j] == s[j - 1]) j += 1;
        ans += j - i - 1;
        i = j - 1;
    }
    cout<<min(ans + 2 * k, n - 1);
    return 0;
}