// APIO 2014 Problem 1 - Palindromes
#include <tuple>
#include <string>
#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
#include <unordered_set>
using namespace std;
int N; unordered_set<int> s1[600009], s2[600009];
tuple<long long, int, int, int> solve(int l, int r, vector<int> &lcp, vector<pair<int, int> > &trait) {
// returns = (answer, store-pos, ct1, ct2)
int baselen = trait[l].second;
for(int i = l; i < r - 1; ++i) {
baselen = min(baselen, lcp[i]);
}
if(r - l == 1) {
if(trait[l].first == 1) s1[l].insert(trait[l].second);
if(trait[l].first == 2) s2[l].insert(trait[l].second);
return make_tuple(1LL, l, 0, 0);
}
long long ans = 0;
vector<tuple<int, int, int> > g;
int pre = l;
for(int i = l; i < r; ++i) {
if(i == r - 1 || lcp[i] == baselen) {
tuple<long long, int, int, int> res = solve(pre, i + 1, lcp, trait);
ans = max(ans, get<0>(res));
pre = i + 1;
g.push_back(make_tuple(get<1>(res), get<2>(res), get<3>(res)));
}
}
int mxpos = -1, mxsize = 0;
for(int i = 0; i < g.size(); ++i) {
int sz = s1[get<0>(g[i])].size() + s2[get<0>(g[i])].size();
if(mxsize < sz) {
mxsize = sz;
mxpos = i;
}
}
int p = get<0>(g[mxpos]);
int ct1 = get<1>(g[mxpos]) * 2, ct2 = get<2>(g[mxpos]) * 2;
for(int i = 0; i < g.size(); ++i) {
if(i == mxpos) continue;
for(int j : s1[get<0>(g[i])]) {
if(s2[p].find(N - j) != s2[p].end()) ++ct1;
if(s2[p].find(N - j + 1) != s2[p].end()) ++ct2;
}
for(int j : s2[get<0>(g[i])]) {
if(s1[p].find(N - j) != s1[p].end()) ++ct1;
if(s1[p].find(N - j + 1) != s1[p].end()) ++ct2;
}
}
for(int i = 0; i < g.size(); ++i) {
if(i == mxpos) continue;
for(int j : s1[get<0>(g[i])]) {
s1[p].insert(j);
}
for(int j : s2[get<0>(g[i])]) {
s2[p].insert(j);
}
}
for(int i = 0; i < g.size(); ++i) {
if(i == mxpos) continue;
for(int j : s1[get<0>(g[i])]) {
if(s2[p].find(N - j) != s2[p].end()) ++ct1;
if(s2[p].find(N - j + 1) != s2[p].end()) ++ct2;
}
for(int j : s2[get<0>(g[i])]) {
if(s1[p].find(N - j) != s1[p].end()) ++ct1;
if(s1[p].find(N - j + 1) != s1[p].end()) ++ct2;
}
}
ct1 /= 2; ct2 /= 2;
ans = max(ans, 1LL * (baselen * 2) * ct1);
ans = max(ans, 1LL * (baselen * 2 - 1) * ct2);
return make_tuple(ans, p, ct1, ct2);
}
int main() {
// step #1. read input
string S;
cin >> S;
N = S.size();
// step #2. construct suffix array of T = S + "#" + rev(S)
string RS = S;
reverse(RS.begin(), RS.end());
string T = S + "#" + RS;
vector<int> sa_inv(2 * N + 1);
for(int i = 0; i < 2 * N + 1; ++i) {
sa_inv[i] = int(T[i]);
}
for(int i = 1; i < 2 * N + 1; i *= 2) {
vector<pair<int, int> > nseq(2 * N + 1);
for(int j = 0; j < 2 * N + 1; ++j) {
nseq[j] = make_pair(sa_inv[j], j + i < 2 * N + 1 ? sa_inv[j + i] : -1);
}
vector<pair<int, int> > sseq(nseq);
sort(sseq.begin(), sseq.end());
sseq.erase(unique(sseq.begin(), sseq.end()), sseq.end());
for(int j = 0; j < 2 * N + 1; ++j) {
sa_inv[j] = lower_bound(sseq.begin(), sseq.end(), nseq[j]) - sseq.begin();
}
}
vector<int> sa(2 * N + 1);
for(int i = 0; i < 2 * N + 1; ++i) {
sa[sa_inv[i]] = i;
}
// step #3. construct rolling-hash table and rolling-hash function
const int mod = 469762049;
const int base = 311;
vector<int> pw(2 * N + 2), h(2 * N + 2);
pw[0] = 1;
for(int i = 0; i < 2 * N + 1; ++i) {
pw[i + 1] = 1LL * pw[i] * base % mod;
h[i + 1] = (1LL * h[i] * base + T[i]) % mod;
}
function<int(int, int)> gethash = [&](int l, int r) {
return (h[r] - 1LL * h[l] * pw[r - l] % mod + mod) % mod;
};
// step #4. calculate LCP
vector<int> lcp(2 * N);
for(int i = 0; i < 2 * N; ++i) {
int l = 0, r = (2 * N + 1) - max(sa[i], sa[i + 1]) + 1;
while(r - l > 1) {
int m = (l + r) >> 1;
if(gethash(sa[i], sa[i] + m) == gethash(sa[i + 1], sa[i + 1] + m)) l = m;
else r = m;
}
lcp[i] = l;
}
// step #5. calculate types and lengths of elements in suffix array
vector<pair<int, int> > trait(2 * N + 1);
for(int i = 0; i < 2 * N + 1; ++i) {
if(sa[i] == N) trait[i] = make_pair(0, -1);
else if(sa[i] < N) trait[i] = make_pair(1, N - sa[i]);
else trait[i] = make_pair(2, (2 * N + 1) - sa[i]);
}
// step #6. calculate the answer
tuple<long long, int, int, int> ans = solve(1, 2 * N + 1, lcp, trait);
// step #7. print the answer
cout << get<0>(ans) << endl;
return 0;
}
