#include <bits/stdc++.h>
using namespace std;
#define int long long
int n;
char a[300005];
int modulo = 1e9 + 123;
int base = 41;
int powbase[300005],invpowbase[300005];
int h[300005],revh[300005];///sum din a[i] * B^(i - 1)
int lgpow(int x,int y)
{
int z = 1;
while (y != 0)
{
if (y % 2 == 1)
z = z * x % modulo;
x = x * x % modulo;
y /= 2;
}
return z;
}
vector<double> ev[300005];///cu + daca il bag, - daca il scot
bool pal(int l,int r)
{
if (l < 1 or r > n or l > r)
return false;
for (int i = l; i <= r; i++)
if (a[i] != a[r + l - i])
return false;
return true;
int h1 = (h[r] - h[l - 1] + modulo) % modulo * invpowbase[l - 1] % modulo;
int h2 = (revh[l] - revh[r + 1] + modulo) % modulo * invpowbase[n - r] % modulo;
if (h1 == h2)
return true;
return false;
}
int get_hash(int l, int r)
{
int h1 = (h[r] - h[l - 1] + modulo) % modulo * invpowbase[l - 1] % modulo;
return h1;
}
int lft[300005];
map<int,int> id;
pair<int,int> what[300005];///sa stiu si eu cam cum arata un string de acel id
int cnt_id;
int f[300005];
int l2[300005];
signed main()
{
string when_what_where_how_why;
cin >> when_what_where_how_why;
n = when_what_where_how_why.size();
for (int i = 1; i <= n; i++)
a[i] = when_what_where_how_why[i - 1];
powbase[0] = invpowbase[0] = 1;
for (int i = 1; i <= n; i++)
{
powbase[i] = base * powbase[i - 1] % modulo;
if (i >= 2)
invpowbase[i] = invpowbase[1] * invpowbase[i - 1] % modulo;
else
invpowbase[i] = lgpow(base,modulo - 2);
}
for (int i = 1; i <= n; i++)
h[i] = (h[i - 1] + (a[i] - 'a' + 1) * powbase[i - 1]) % modulo;
for (int i = n; i >= 1; i--)
revh[i] = (revh[i + 1] + (a[i] - 'a' + 1) * powbase[n - i]) % modulo;
for (int i = 1; i <= n; i++)
{
int st = 0,pas = 1 << 18;
while (pas != 0)
{
if (pal(i - st - pas,i + st + pas))
st += pas;
pas /= 2;
}
ev[i].push_back(i);
ev[i + st + 1].push_back(-i);
}
for (int i = 1; i < n; i++)
{
int st = 0,pas = 1 << 18;
while (pas != 0)
{
if (pal(i - st - pas + 1,i + st + pas))
st += pas;
pas /= 2;
}
if (st == 0)
continue;
double lol = (double)i + 0.5d;
ev[i + 1].push_back(lol);
ev[i + st + 1].push_back(-lol);
}
multiset<double> ms;
for (int i = 1; i <= n; i++)
{
for (auto it : ev[i])
{
double vl = it;
if (vl > 0)
ms.insert(vl);
else
ms.erase(ms.find(-vl));
}
double hmm = *ms.begin();
double d = (double)i - hmm;
d *= (2.0d);
lft[i] = i - d;
if (ms.size() == 1)
continue;
ms.erase(ms.find(hmm));
double hmm2 = *ms.begin();
d = (double)i - hmm2;
d *= (2.0d);
l2[i] = i - d;
ms.insert(hmm);
}
for (int i = 1; i <= n; i++)
{
int cur = get_hash(lft[i],i);
if (!id[cur])
id[cur] = ++cnt_id;
what[id[cur]] = {lft[i],i};
f[id[cur]]++;
}
vector<pair<int,int>> ord;
for (int i = 1; i <= cnt_id; i++)
ord.push_back({what[i].second - what[i].first + 1,i});
sort(ord.begin(),ord.end());
reverse(ord.begin(),ord.end());
int ans = 0;
for (auto it : ord)
{
ans = max(ans,it.first * f[it.second]);
if (it.first == 1)
continue;
int h2 = get_hash(l2[it.second],it.second);
int id2 = id[h2];
f[id2] += f[it.second];
}
cout << ans;
return 0;
}
/**
Maxim N palindroame distincte etc
Voi vrea pentru fiecare sa retin frecventa lui
Pentru asta, merg cu i de la 1 la N, iau cel mai mare palindrom care se termina pe i
Ii gasesc id-ul (ori un id nou ori ceva id vechi fiindca deja exista)
Acum, vreau sa adaug 1 pe toate palindroamele care se termina pe i <=> toate sufixele palindrom ale lui pal[id] <=> toate prefixele palindrom
Dau f[id]++
La final, vreau sa propag f-urile pe toate prefixele palindrom
Foarte simplu, iau palindroamele descrescator dupa lungime, pentru asta o sa am f[id] updatat si il consider la raspuns
Fie pal[id'] = prefixul maxim palindrom al lui pal[id] (daca pal[id] are lungime 1, ma opresc)
Dau f[id'] += f[id] and we move on
**/
