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 = 2e6 + 5;
const int inf = 1e9 + 5;
const double pi = acos(-1);
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};
const int mod = 1e9 + 7;
int p[N], cnt[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  string s;
  cin >> s;
  int n = (int)(s).size();
  memset(p, 0, sizeof p);
  for (int i = 1; i < n; i++) {
    int prev = p[i - 1];
    while (prev > 0 && s[prev] != s[i]) prev = p[prev - 1];
    if (s[prev] == s[i]) prev++;
    p[i] = prev;
  }
  if (p[n - 1] == 0) {
    if (p[p[n - 1] - 1] == 0) {
      cout << "Just a legend";
    } else {
      cout << s.substr(0, p[p[n - 1] - 1]);
    }
  } else {
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      if (p[i] == p[n - 1]) {
        ++cnt;
      }
    }
    if (cnt >= 2) {
      cout << s.substr(0, p[n - 1]);
    } else {
      if (p[p[n - 1] - 1] == 0) {
        cout << "Just a legend";
      } else {
        cout << s.substr(0, p[p[n - 1] - 1]);
      }
    }
  }
  return 0;
}