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 N = 1e6 + 6;
class cf126b {
 public:
  int p[N];
  void solve(std::istream& in, std::ostream& out) {
    string s;
    in >> s;
    int n = s.length();
    kmp(s);
    int x = 0, ans = p[n - 1];
    for (int i = 1; i < n - 1; i++) x = max(x, p[i]);
    if (ans == 0)
      out << "Just a legend";
    else {
      while (ans) {
        if (ans <= x) break;
        ans = p[ans - 1];
      }
      if (!ans)
        out << "Just a legend";
      else
        for (int i = 0; i < ans; i++) out << s[i];
    }
  }
  void kmp(string s) {
    int n = s.length();
    memset(p, 0, sizeof(p));
    for (int i = 1; i < n; i++) {
      int j = p[i - 1];
      while (j && s[j] != s[i]) j = p[j - 1];
      if (s[i] == s[j]) j++;
      p[i] = j;
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cf126b solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}