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>
const int N = 1000009;
const int p = 31;
const int m = 1e9 + 7;
using namespace std;
int n;
string s;
long long p_pow[N], h[N + 1];
void preProcess(string const& s) {
  p_pow[0] = 1;
  h[0] = 0;
  n = s.size();
  for (int i = 1; i <= n; i++) p_pow[i] = (p_pow[i - 1] * p) % m;
  for (int i = 1; i < n; i++)
    h[i] = (h[i - 1] + (s[i] - 'a' + 1) * p_pow[i]) % m;
}
bool countHash(int i, int l) {
  long long cur_h = (h[l] + m - h[0]) % m;
  cur_h = (cur_h * p_pow[i - 1]) % m;
  long long le = (h[i + l - 1] + m - h[i - 1]) % m;
  return cur_h == le;
}
int computeSuffix(int l, int r) { return (h[r] + m - h[l - 1]) % m; }
long long check(int i) {
  int l = 1, r = n - i - 1, res = -1;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (countHash(i, mid)) {
      res = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  if (res == -1) return -1;
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  cin >> s;
  s = '#' + s;
  n = s.length() - 1;
  preProcess(s);
  pair<int, int> res = {-1, -1};
  for (int i = 2; i < n; i++) {
    long long t = check(i);
    if (res.second < t) {
      res = {i, t};
    }
  }
  int ma = res.second, le = -1;
  n = s.length() - 1;
  for (int i = 1; i < s.length() - 1; i++)
    if ((computeSuffix(1, i) * p_pow[n - i]) % m ==
            computeSuffix(n - i + 1, n) &&
        i <= ma) {
      le = i;
    }
  int ss = 1;
  if (le != -1)
    while (le-- > 0) cout << s[ss++];
  else
    cout << "Just a legend";
}