#include <bits/stdc++.h>
using namespace std;
#define int long long
const int INF = 1e18;
const int MAXN = 1e6 + 5;
typedef long double ld;
struct line
{
int m, b;
int eval(int x) {return m * x + b;}
};
ld interx(line a, line b)
{
return ((ld)b.b - (ld)a.b) / ((ld)a.m - (ld)b.m);
}
int n,k;
string s;
int pozb[MAXN];
int dp[MAXN], cnt_songs[MAXN];
int auxdp[MAXN];
pair<int, int> solve(int coef)//{{rez_coef, {nr_swaps, cnt_songs}}
{
/*dp[0] = 0;
cnt_songs[0] = 0;
int lim = 0;
int balance = 0;
for(int i=1;i<=n;i++)
{
while(lim + 1 <= i && pozb[lim + 1] < i)
lim++;
dp[i] = INF;
cnt_songs[i] = -1;
auxdp[i-1] = dp[i-1] + coef - balance;
balance += max(0LL, (i-1) - pozb[i]);
for(int x=0;x<lim;x++)
auxdp[x] += lim - x;
for(int x=i-1;x>=0;x--)
{
if(auxdp[x] + balance < dp[i])
{
dp[i] = auxdp[x] + balance;
cnt_songs[i] = cnt_songs[x] + 1;
}
}
}*/
int lim = 0, balance = 0, balance_hull = 0;
deque<line> hull;
dp[0] = 0;
cnt_songs[0] = 0;
for(int i=1;i<=n;i++)
{
dp[i] = INF;
cnt_songs[i] = -1;
auxdp[i-1] = dp[i-1] + coef - balance;
balance += max(0LL, (i-1) - pozb[i]);
while(lim + 1 <= i && pozb[lim + 1] < i)
{
hull.push_back({-lim, auxdp[lim] + lim * (i-1) - balance_hull});
lim++;
}
//for(int x=0;x<lim;x++)
// auxdp[x] += lim - x;
balance_hull += lim;
for(int x=i-1;x>=lim;x--)
{
if(auxdp[x] + balance < dp[i])
{
dp[i] = auxdp[x] + balance;
cnt_songs[i] = cnt_songs[x] + 1;
}
}
for(int u=(int)hull.size()-1;u>=0;u--)
{
line l = hull[u];
int cv = l.eval(i) + balance + balance_hull;
if(cv < dp[i])
{
dp[i] = cv;
cnt_songs[i] = cnt_songs[-l.m] + 1;
}
}
}
return {dp[n], cnt_songs[n]};
}
signed main()
{
ios_base::sync_with_stdio(0);cin.tie(0);
cin>>n>>k>>s;
assert(s.size() == 2 * n);
int cnta = 0, cntb = 0;
for(int i=0;i<2*n;i++)
{
if(s[i] == 'A')
{
cnta++;
}
else
{
cntb++;
pozb[cntb] = cnta;
}
}
assert(cnta == n && cntb == n);
int st = 0, dr = 1e9, ans = -1;
while(st <= dr)
{
int mij = (st + dr) / 2;
auto x = solve(mij);
if(x.second >= k)
{
ans = mij;
st = mij + 1;
}
else
{
dr = mij - 1;
}
}
assert(ans != -1);
auto x = solve(ans);
cout<<x.first - k * ans;
return 0;
}
