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;
string ss;
int pf[1000005];
int mxp = 0;
void prefix(string &s, int p[]) {
  int i, k = 0;
  p[0] = 0;
  for (i = 1; i < s.size(); i++) {
    while (k && s[i] != s[k]) k = p[k - 1];
    if (s[i] == s[k]) k++;
    p[i] = k;
    if (i != s.size() - 1) mxp = max(mxp, p[i]);
  }
}
int main() {
  int i, j, k;
  cin >> ss;
  prefix(ss, pf);
  k = pf[ss.size() - 1];
  while (k > mxp && k) k = pf[k - 1];
  if (k == 0)
    cout << "Just a legend\n";
  else
    for (i = 0; i < k; i++) cout << ss[i];
  return 0;
}