/*
 * Rezolvarea naiva in N^3.
 *
 * Se putea si mai rapid in N^2 folosind doua cozi. In prima bagam primul caracter si in a doua urmatorul.
 * Apoi pentru a creste lungimea bagam in prima coada primul caracter din a doua coada, iar aceasta
 * primea inca 2 caractere (daca mai existau). La fiecare scoatere-bagare se actualizau doua numere
 * pe post de hashuri. Aceste doua hashuri se comparau in O(1) folosind `==` astfel scutind inca o dimensiune.
 *
 * Calcularea numerelor se facea astfel: se scadea din caracterul curent caracterul 'A' ('A' < 'a') si se aduna la suma.
 * Inainte de adunare suma era inmultita cu 100 (pentru evitarea coliziunilor evidente) pentru prima suma (specifica primei cozi).
 * Din a doua suma se scadea valoarea primului caracter si se adunau pe rand primul caracter la puterea `p` si al doilea la puterea `p` + 1.
 * Unde `p` reprezinta numarul de caractere din prima coada - 1 (adica jumatate de palindrom).
 *
 * Deoarece sumele sunt foarte mari se facea modulo cu o constanta destul de mare si pentru evitarea coliziunilor mai putin evidente
 * se foloseau vreo `k` constante (3-6) diferite pentru a compara sumele intre ele, astfel crescand acuratetea.
 * Complexitatea finala devenea O(KN^2).
 */

#include <iostream>
#include <string>
using namespace std;


int maxPal; // dimensiunea maxima
string str, pal; // sirul sursa, palindromul maxim


void solve()
{
    /**
     * Pentru fiecare pozitie si lungime se genereaza o subsecventa
     * ce se compara cu ea insasi pornind din extremitati.
     * Probabil daca se genereaza si inversul acesteia dureaza ceva mai mult
     * chiar daca complexitatea teoretica ramane aceeasi.
     */
    for (int pos = 0; pos < str.size(); ++pos) {
        for (int len = 1; len <= str.size() - pos; ++len) {
            string sub = str.substr(pos, len);
            bool isPal = true;
            for (int i = 0; i < len / 2 && isPal; i++) {
                if (sub[i] != sub[len - i - 1]) isPal = false;
            }
            if (isPal && len > maxPal) {
                maxPal = len;
                pal = sub;
            }
        }
    }
}


int main()
{
    cin >> str;
    solve(); // rezolvam
    cout << pal;
    return 0;
}

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