#include <bits/stdc++.h>
using namespace std;
#define int long long
const int INF = 1e18;
const int MAXN = 5e3 + 5;
int n,k;
string s;
int pozb[MAXN];
typedef long long ld;
pair<ld, pair<int,int>> solve(ld coef)//{{rez_coef, {nr_swaps, cnt_songs}}
{
vector<ld> dp(n+2);
vector<int> real_dp(n+2), cnt_songs(n+2);
vector<int> pref(n+2,0);
dp[0] = 0;
real_dp[0] = 0;
cnt_songs[0] = 0;
for(int i=1;i<=n;i++)
{
dp[i] = INF;
real_dp[i] = INF;
cnt_songs[i] = -1;
int aux = 0;
for(int old=i-1;old>=0;old--)
{
aux += max(0LL, i - pozb[old + 1]);
if(real_dp[old] >= INF)
continue;
ld cv = dp[old] + aux + coef;
if(cv < dp[i])
{
dp[i] = cv;
real_dp[i] = real_dp[old] + aux;
cnt_songs[i] = cnt_songs[old] + 1;
}
}
}
return {dp[n], {real_dp[n], cnt_songs[n]}};
}
void easy_dp()
{
vector<vector<int>> dp(n+2, vector<int>(k+2,INF));
vector<int> pref(n+2,0);
for(int i=0;i<=n;i++)
for(int cnt=0;cnt<=k;cnt++)
dp[i][cnt] = INF;
dp[0][0] = 0;
for(int i=1;i<=n;i++)
{
pref[0] = 0;
for(int u=1;u<=n;u++)
pref[u] = pref[u-1] + max(0LL, i - pozb[u]);
for(int cnt=1;cnt<=k;cnt++)
{
int mnm = INF;
for(int old=0;old<i;old++)
mnm = min(mnm, dp[old][cnt-1] - pref[old]);
dp[i][cnt] = min(INF, mnm + pref[i]);
}
}
cout<<dp[n][k];
}
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;
}
}
//for(int i=1;i<=n;i++)cerr<<pozb[i]<<" ";cerr<<"pozb\n";
assert(cnta == n && cntb == n);
/*int rez = INF;
ld st = 0, dr = 1e9;
for(int pas=0;pas<100;pas++)
{
//the greater coef is, the greater cost_song is, so the less songs there are
ld mij = (st + dr) / 2;
auto x = solve(mij);
if(x.second.second == k) rez = min(rez, x.second.first);
if(x.second.second <= k)
dr = mij;
else
st = mij;
}
cout<<rez;*/
int st = 0, dr = 1e9, ans = -1;
while(st <= dr)
{
int mij = (st + dr) / 2;
auto x = solve(mij);
if(x.second.second <= k)
{
ans = mij;
dr = mij - 1;
}
else
{
st = mij + 1;
}
}
assert(ans != -1);
auto x = solve(ans);
cout<<x.first - k * ans;
return 0;
}
