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 INF = 1e9 + 9;
const long long LINF = 1e12;
const int MAXN = 1e6 + 5;
const int MOD = 1e9 + 7;
const double eps = 1e-14;
int n;
int pf[MAXN];
string s;
void pref_func() {
  for (int i = 1; i < n; i++) {
    int j = pf[i - 1];
    while (j > 0 && s[i] != s[j]) j = pf[j - 1];
    if (s[i] == s[j]) j++;
    pf[i] = j;
  }
}
void solve() {
  cin >> s;
  n = s.size();
  pref_func();
  int k = 0;
  for (int i = 1; i < n - 1; i++) {
    if (pf[i] == pf[n - 1]) k = pf[n - 1];
  }
  for (int i = 1; i < n - 1; i++) {
    if (pf[i] == pf[pf[n - 1] - 1]) k = max(k, pf[pf[n - 1] - 1]);
  }
  if (k) {
    for (int i = 0; i < k; i++) cout << s[i];
  } else
    cout << "Just a legend";
}
int main() {
  if (!1) {
    freopen(
        "474"
        ".in",
        "r", stdin);
    freopen(
        "474"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}