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;
string a;
bool kmp(string s) {
  vector<int> v;
  v.resize(s.size());
  v[0] = 0;
  for (int i = 1, j = 0; i < s.size(); ++i) {
    if (s[i] == s[j]) {
      v[i] = j + 1;
      ++j;
    } else {
      if (j == 0)
        ;
      else {
        j = v[j - 1];
        --i;
      }
    }
  }
  string str = s.substr(0, v[v.size() - 1]);
  v.resize(str.size());
  v[0] = 0;
  for (int i = 1, j = 0; i < str.size(); ++i) {
    if (str[i] == str[j]) {
      v[i] = j + 1;
      ++j;
    } else {
      if (j == 0)
        ;
      else {
        j = v[j - 1];
        --i;
      }
    }
  }
  if (str.size() == 0) return false;
  int k = 0;
  int ans = 0;
  for (int i = 0; i < a.size(); ++i) {
    if (a[i] == str[k]) {
      ++k;
      if (k == str.size()) {
        ++ans;
        if (k > 0) k = v[k - 1];
      }
    } else {
      if (k > 0) {
        k = v[k - 1];
        --i;
      }
    }
  }
  if (ans > 2) {
    cout << str;
    return true;
  }
  return kmp(str);
}
int main() {
  cin >> a;
  if (!kmp(a)) cout << "Just a legend";
}