#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define st first
#define nd second
typedef long long ll;
typedef long double ld;
const ll I = 1'000'000'000'000'000'000LL;
const int II = 2'000'000'000;
const ll M = 1'000'000'007LL;
const int N = 1<<20;
int tab[N];
pair<ll, int> dp[N];
pair<ll, ll> f[N];
vector<int> cur;
int pos = 0; ll dod = 0LL;
pair<ll, int> Get(int a, ll v)
{
return pair{f[a].st * v + f[a].nd + dod, dp[a].nd};
}
ll It(int a, int b)
{
ll d = f[a].nd - f[b].nd;
ll r = f[b].st - f[a].st;
ll p = (d + r - 1LL) / r;
if(Get(a, p - 1) <= Get(b, p - 1))
return p - 1;
if(Get(a, p) > Get(b, p))
return p + 1;
return p;
}
void Add(int i, int n, ll t)
{
f[i] = pair{(ll)(n - i), dp[i].st - (ll)(n - i) * t};
f[i].nd -= dod;
int m = (int)cur.size() - 1;
//cout << "\nA: " << i << " : " << f[i].st << " " << f[i].nd << "\n";
while(m >= pos + 1 && It(i, cur[m]) <= It(cur[m], cur[m - 1]))
{
//cout << "D: " << cur.back() << "\n";
--m;
cur.pop_back();
}
//cout << "\n";
cur.pb(i);
}
void Licz(int n, ll x)
{
cur.clear();
multiset<pair<ll, int>> akt;
akt.insert(dp[0]);
int il = 0, j = -1;
pos = 0; dod = 0LL;
for(int i = 1; i <= n; ++i)
{
dp[i] = pair{I, II};
il += tab[i] - tab[i - 1] - 1;
//cerr << "\nL: " << i << " " << il << '\n';
while(j < min(il, i - 1))
{
++j;
akt.erase(akt.find(dp[j]));
Add(j, n, i - 1);
}
dod += -(n - il);
while(pos < (int)cur.size() - 1 && Get(cur[pos], i) >= Get(cur[pos + 1], i))
++pos;
if((int)akt.size() > 0)
dp[i] = *akt.begin();
if(pos < (int)cur.size())
dp[i] = min(dp[i], Get(cur[pos], i));
--dp[i].nd; dp[i].st += x;
//cout << cur[pos] << "\n";
//cout << "DP: " << dp[i].st << " " << dp[i].nd << "\n";
akt.insert(dp[i]);
}
}
ll BS(int n, int x)
{
ll p = 0, k = 1'000'000'000'000LL, s;
while(p < k)
{
s = (p + k + 1) / 2;
//cerr << p << " " << s << " " << k << "\n";
Licz(n, s);
if(-dp[n].nd >= x)
p = s;
else
k = s - 1;
}
return p;
}
void Solve()
{
int n, k, cnt = 0;
string s;
cin >> n >> k;
cin >> s;
for(int i = 1; i <= 2 * n; ++i)
if(s[i - 1] == 'A')
tab[++cnt] = i;
ll x = 4LL, ans = 0LL;
x = BS(n, k);
//cout << x << "\n";
Licz(n, x);
ans = dp[n].st - (ll)k * x;
cout << ans << "\n";
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
//int t; cin >> t;
//while(t--)
Solve();
return 0;
}
