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;
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  string s;
  cin >> s;
  int n = s.length();
  int i, j, l = 0, r = 0;
  vector<int> z(n, 0);
  for (i = 1; i < n; i++) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (z[i] + i < n && s[z[i]] == s[z[i] + i]) z[i]++;
    if (i + z[i] - 1 > r) {
      l = i;
      r = i + z[i] - 1;
    }
  }
  int mx[n], ans = -1;
  mx[0] = z[0];
  for (i = 1; i < n; i++) mx[i] = max(mx[i - 1], z[i]);
  for (i = n - 1; i > 0; i--) {
    if (i + z[i] == n && mx[i - 1] >= z[i]) ans = i;
  }
  if (ans == -1)
    cout << "Just a legend\n";
  else {
    for (int j = ans; j < n; j++) cout << s[j];
    cout << "\n";
  }
  return 0;
}