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;
char a[1000005];
int ne[1000005], n, maxx = 0;
void c() {
  ne[1] = 0;
  for (int i = 2, j = 0; i <= n; i++) {
    while (j && a[i] != a[j + 1]) {
      j = ne[j];
    }
    if (a[i] == a[j + 1]) {
      j++;
    }
    ne[i] = j;
    if (i != n) {
      maxx = max(ne[i], maxx);
    }
  }
}
int main() {
  cin >> a + 1;
  n = strlen(a + 1);
  c();
  int x = ne[n];
  if (x == 0) {
    printf("Just a legend\n");
  } else {
    while (x > maxx) {
      x = ne[x];
    }
    if (x == 0) {
      printf("Just a legend\n");
      return 0;
    }
    for (int i = 2; i < n; i++)
      if (x == ne[i]) {
        for (int j = i - ne[i] + 1; j <= i; j++) {
          printf("%c", a[j]);
        }
        printf("\n");
        return 0;
      }
  }
  return 0;
}