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;
long long t = 1, n, xx, x = 0, y = 0, cc, f, m, k, base = 397;
long long hsh[2000005], ans;
long long pb[2000000 + 5];
string s, ss = "";
long long get_hash(string str, int sz) {
  long long ans = 0;
  for (int i = 0; i < sz; i++) {
    ans += ((str[sz - i - 1] - 'a' + 1) * pb[i]) % 1000000009;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  s = "F" + s;
  n = s.size();
  pb[0] = 1;
  for (int i = 1; i < 1000001; i++) {
    pb[i] = (pb[i - 1] * base) % 1000000009;
  }
  hsh[0] = 0ll;
  for (int i = 1; i <= n; i++) {
    hsh[i] = ((hsh[i - 1] * base) % 1000000009 + (s[i] - 'a' + 1)) % 1000000009;
  }
  long long val, vall;
  f = 0;
  for (int i = n - 3; i > 0; i--) {
    val = hsh[i];
    vall = (hsh[n - 1] - ((hsh[n - i - 1] * pb[i]) % 1000000009) + 1000000009) %
           1000000009;
    if (val == vall) {
      for (int j = 2; j < n - i; j++) {
        vall = (hsh[j + i - 1] - ((hsh[j - 1] * pb[i]) % 1000000009) +
                1000000009) %
               1000000009;
        if (val == vall) {
          f = 1;
          ans = i;
          break;
        }
      }
      if (f) break;
    }
  }
  if (!f)
    cout << "Just a legend" << endl;
  else {
    for (int i = 1; i < ans + 1; i++) cout << s[i];
    cout << endl;
  }
  return 0;
}