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;
const int nn = 1100000;
int f[nn], L;
char S[nn];
int main() {
  scanf("%s", S + 1);
  L = strlen(S + 1);
  while (f[2] < L - 1 && S[1 + f[2]] == S[2 + f[2]]) f[2]++;
  int l = 2, r = l + f[2] - 1;
  for (int i = 3; i <= (int)(L); ++i) {
    if (i > r) {
      l = i;
      r = i - 1;
      while (r < L && S[r + 1] == S[r + 1 - l + 1]) r++;
      f[i] = r - l + 1;
    } else {
      if (i + f[i - l + 1] - 1 < r) {
        f[i] = f[i - l + 1];
        continue;
      }
      l = i;
      while (r < L && S[r + 1] == S[r + 1 - l + 1]) r++;
      f[i] = r - l + 1;
    }
  }
  int mx = 0, x = 0;
  for (int i = 2; i <= (int)(L); ++i) {
    if (i + f[i] - 1 == L && i + mx > L) {
      x = L - i + 1;
      break;
    }
    mx = max(mx, f[i]);
  }
  if (x) {
    for (int i = 1; i <= (int)(x); ++i) putchar(S[i]);
    puts("");
  } else
    puts("Just a legend");
  return 0;
}