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;
void suffix(char *s, int *z) {
  int l = 0, r = 0, i;
  for (i = 1; s[i]; i++) {
    if (r <= i) {
      for (r = i; s[r] == s[r - i]; r++)
        ;
      l = i;
      z[i] = r - l;
    } else {
      if (z[i - l] < r - i)
        z[i] = z[i - l];
      else {
        z[i] = r - i;
        while (s[z[i]] == s[r]) z[i]++, r++;
        l = i;
      }
    }
  }
}
char s[1024 * 1024];
int z[1024 * 1024];
int main() {
  gets(s);
  suffix(s, z);
  int n = strlen(s);
  int i;
  int mx = -1;
  for (i = 1; s[i]; i++)
    if ((z[i] > mx && i + z[i] < n) || z[i] - 1 > mx) {
      mx = z[i];
      if (i + z[i] == n) mx--;
    }
  for (i = 1; s[i] && !(z[i] + i == n && z[i] <= mx); i++)
    ;
  if (i == n)
    puts("Just a legend");
  else
    puts(s + i);
  fclose(stdin);
  fclose(stdout);
  return 0;
}