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 n, i, k;
string s1, s2;
int F[1000010];
void kmp(string s) {
  for (i = 1; i < s.size(); i++) {
    while (s[i] != s[k] && k > 0) {
      k = F[k - 1];
    }
    if (s[i] == s[k]) k++;
    F[i] = k;
  }
}
int main() {
  cin >> s1;
  n = s1.size();
  kmp(s1);
  if (F[n - 1] == 0) return cout << "Just a legend", 0;
  for (int i = 0; i < n - 1; i++) {
    if (F[i] == F[n - 1]) {
      for (int j = 0; j < F[i]; j++) cout << s1[j];
      return 0;
    }
  }
  if (F[F[n - 1] - 1] == 0) return cout << "Just a legend", 0;
  for (int j = 0; j < F[F[n - 1] - 1]; j++) cout << s1[j];
  return 0;
}