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>
using namespace std;
int main() {
  string pattern;
  cin >> pattern;
  long long int i, m, n, j;
  n = pattern.length();
  long long int dfa[n];
  memset(dfa, 0, sizeof(dfa));
  long long int x = 0;
  for (i = 1; i < n; i++) {
    if (pattern[i] == pattern[x])
      x++, dfa[i] = x;
    else if (x > 0) {
      x = dfa[x - 1];
      i--;
    }
  }
  for (i = 0; i < n - 1; i++)
    if (dfa[n - 1] > 0 && dfa[i] == dfa[n - 1]) {
      for (j = 0; j < dfa[n - 1]; j++) printf("%c", pattern[j]);
      cout << endl;
      return 0;
    }
  if (i == n - 1 && dfa[n - 1] > 0 && dfa[dfa[n - 1] - 1] > 0) {
    for (j = 0; j < dfa[dfa[n - 1] - 1]; j++) printf("%c", pattern[j]);
    cout << endl;
    return 0;
  } else if (i == n - 1)
    cout << "Just a legend\n";
  return 0;
}