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 pref[1000005], pot[1000005], inv[1000005];
int mod = 1e9 + 9;
string s;
int modpow(int b, int e) {
  if (e == 0) return 1;
  if (e == 1) return b;
  int res = modpow(b, e / 2);
  res = ((long long)res * res) % mod;
  if (e % 2 == 1) res = ((long long)res * b) % mod;
  return res;
}
int main() {
  cin >> s;
  int n = s.size();
  s = "$" + s;
  pot[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++) {
    pot[i] = ((long long)pot[i - 1] * 7919) % mod;
    inv[i] = modpow(pot[i], mod - 2);
    pref[i] = (pref[i - 1] + (s[i] - 'a' + 1LL) * pot[i]) % mod;
  }
  for (int i = 2; i <= n; i++) {
    int sum = ((long long)pref[n] - pref[i - 1] + mod) % mod;
    sum = ((long long)sum * inv[i - 1]) % mod;
    int othersum = pref[n - i + 1];
    if (sum == othersum) {
      int len = n - i + 1;
      for (int j = 2; j + len - 1 < n; j++) {
        int sum2 = ((long long)pref[j + len - 1] - pref[j - 1] + mod) % mod;
        sum2 = ((long long)sum2 * inv[j - 1]) % mod;
        if (sum2 == sum) {
          for (int k = 1; k <= len; k++) printf("%c", s[k]);
          printf("\n");
          return 0;
        }
      }
    }
  }
  puts("Just a legend");
  return 0;
}