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 p[2000000];
bool used[2000000];
string s;
int main(void) {
  cin >> s;
  p[0] = 0;
  int n = s.length();
  for (int i = 1; i < n; i++) {
    int j = p[i - 1];
    while (j && s[j] != s[i]) j = p[j - 1];
    if (s[j] == s[i]) p[i] = j + 1;
  }
  for (int i = 1; i < n - 1; i++) {
    used[p[i]] = 1;
  }
  int i = p[n - 1];
  while (i && !used[i]) i = p[i - 1];
  if (i)
    cout << s.substr(0, i);
  else
    cout << "Just a legend";
  return 0;
}