// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

#define    flash           ios_base::sync_with_stdio(false);cin.tie(0);

#define    ff              first

#define    ss              second

#define    pb              push_back

#define    m_p             make_pair

#define    ret             return 0

#define    MAX(a,b)        max({a,b})

#define    MAX(a,b,c)      max({a,b,c})

#define    MAX(a,b,c,d)    max({a,b,c,d})

#define    MIN(a,b)        min({a,b})

#define    MIN(a,b,c)      min({a,b,c})

#define    MIN(a,b,c,d)    min({a,b,c,d})

#define    GCD(a,b)        __gcd(a,b)

#define    LCM(a,b)        (a*b)/GCD(a,b)

#define    MEM(a,b)        memset(a,b,sizeof a)

#define    SC(a)           scanf("%d",&a)

#define    SC2(a,b)        scanf("%d %d",&a,&b)

#define    SC3(a,b,c)      scanf("%d %d %d",&a,&b,&c)

#define    SC4(a,b,c,d)    scanf("%d %d %d %d",&a,&b,&c,&d)

#define    PCs(a)          printf("Case %d: ",a)

#define    WRITE(a)        freopen(a,"w",stdout)

#define    READ(a)         freopen(a,"r",stdin)

#define    LB(a,x)         (lower_bound(a.begin(),a.end(),x) - a.begin())

#define    UB(a,x)         (upper_bound(a.begin(),a.end(),x) - a.begin())

#define    PI              2.0*acos(0.0)

#define    MOD1            1000000007 // prime

#define    MOD2            1000000009 // prime

#define    MOD3            1000000021 // prime

#define    Base1           10000019

#define    Base2           10000079

#define    Base3           10000103

//#define    MAX             1000007

typedef    pair<int,int> pii;

typedef    pair<int,pii> ppi;

typedef    pair<pii,int> pip;

typedef    long long int ll;

typedef    unsigned long long int ull;

typedef    pair<ll,ll> pll;

typedef    vector<int> VI;

typedef    vector<pii> Vii;

typedef    vector<VI> VVI; // 2D

//typedef    priority_queue<int> PQ; // MaxHeap

typedef    priority_queue<int,VI,greater<int> > PQ; // MinHeap

/*

inline int StringToInt(string a)

{

    int num;

    stringstream aw(a);

    aw>>num;

    return num;

}*/

/*

inline ll StringToLL(string a)

{

    ll num;

    stringstream aw(a);

    aw>>num;

    return num;

}*/

/*

inline int iPOW(int a,int e)

{

    int num = 1;

    while(e)

    {

        if(e%2)

        {

            num = num * a;

        }

        e/=2;

        a = a * a;

    }

    return num;

}

inline ll LPOW(ll a,ll e)

{

    ll num = 1;

    while(e)

    {

        if(e%2)

        {

            num = num * a;

        }

        e/=2;

        a = a * a;

    }

    return num;

}*/

/*

inline ll BigMod(ll a,ll e,ll mod)

{

    ll num = 1;

    while(e)

    {

        if(e%2)

        {

            a%=mod;

            num%=mod;

            num = num * a;

            num%=mod;

        }

        e/=2;

        a%=mod;

        a = a * a;

        a%=mod;

    }

    return num%mod;

}*/

/*inline ll modInverse(ll A,ll P)

{

    return BigMod(A,P-2,P);

}

ll fac[MAX];

inline void factorial(int n,int mod)

{

    fac[0] = 1;

    fac[1] = 1;

    for(int i=2;i<=MAX;i++) fac[i] = ( (fac[i-1]%mod)*i)%mod;

}

inline int nCr(int n,int r,int mod) // ncr with mod

{

    return ((fac[n] * modInverse(fac[r],mod)%mod)%mod * (modInverse(fac[n-r],mod)%mod) ) %mod;

}

*/

/*

bool isprime[MAX+1000];

int Primes[MAX],id; // for <=10^6

void sieve()

{

    Primes[0] = 2;

    id++;

    for(int i=4;i<=MAX+100;i+=2) isprime[i] = true; // is not a prime

    for(int i=3;i<=MAX+100;i+=2)

    {

        if(isprime[i]==false)

        {

            Primes[id++] = i;

            if(i+i<=MAX)

            for(int j=i+i;j<=100+MAX;j+=i)

                isprime[j] = true;

        }

    }

}*/

//int Dx[]  = {-1,0, 0,1};

//int Dy[]  = { 0,1,-1,0};

//int Dx8[] = {-1,-1,-1,0,1,1, 1, 0};

//int Dy8[] = {-1, 0, 1,1,1,0,-1,-1};

//int Kx[]  = {2,1,-1,-2,-2,-1, 1, 2};

//int Ky[]  = {1,2, 2, 1,-1,-2,-2,-1};

const int MAX=1e5+10,MOD=1e9+7;

int main ()

{

    //flash;

    int N,K,happy=0;cin >> N >> K;

    string S;cin>>S;

    for(int i=0;i<N-1;i++)happy+=(S[i]==S[i+1]);

    cout<<min(N-1,happy+2*K);

    return 0;

}

/*

*/




// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>



using std::min;



const int MAXN=100005;



int main()

{

    int N,K,diffs=0;

    char str[MAXN];



    scanf("%d%d%s",&N,&K,str);



    for (int i=1;i<N;++i)

        if (str[i]!=str[i-1]) diffs++;

    printf("%d",min(N-1-diffs+2*K,N-1));



    return 0;

}


// Fastest known solution for this problem: