Problem: Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.

A little later they found a string s, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring t of the string s.

Prefix supposed that the substring t is the beginning of the string s; Suffix supposed that the substring t should be the end of the string s; and Obelix supposed that t should be located somewhere inside the string s, that is, t is neither its beginning, nor its end.

Asterix chose the substring t so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring t aloud, the temple doors opened. 

You know the string s. Find the substring t or determine that such substring does not exist and all that's been written above is just a nice legend.

Input

You are given the string s whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.

Output

Print the string t. If a suitable t string does not exist, then print "Just a legend" without the quotes.

Examples

Input

fixprefixsuffix


Output

fix

Input

abcdabc


Output

Just a legend

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
const long double PI = acos(-1.0);
const int INF = 1e9;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
const long long BIGINF = 1e18;
mt19937 rng(time(0));
uniform_int_distribution<int> uid(-1000000000, 1000000000);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  vector<int> p(s.size(), 0);
  int l = 0, r = 0;
  p[0] = 0;
  for (int i = 1; i < s.size(); ++i) {
    p[i] = p[i - 1];
    while (p[i] > 0 && s[p[i]] != s[i]) p[i] = p[p[i] - 1];
    if (s[p[i]] == s[i]) p[i]++;
  }
  if (!p.back()) {
    cout << "Just a legend" << '\n';
    return 0;
  }
  int k = p.back();
  while (k > 0) {
    for (int i = 0; i < int(s.size()) - 1; ++i) {
      if (p[i] == k) {
        for (int j = 0; j < k; ++j) {
          cout << s[j];
        }
        cout << '\n';
        return 0;
      }
    }
    k = p[k - 1];
  }
  cout << "Just a legend" << '\n';
  return 0;
}