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;
char s[1000005];
const int p = 31, m = 1e9 + 9;
bool poss(int n, long long hash, int len, vector<long long>& power,
          vector<long long>& h_s) {
  for (int i = 1; i + len - 1 < n - 1; ++i) {
    long long cur = (h_s[i + len] - h_s[i] + m) % m;
    cur = cur * power[n - i - 1] % m;
    if (cur == hash) return true;
  }
  return false;
}
void solve(int n) {
  vector<long long> power(n), h_s(n + 1, 0);
  power[0] = 1;
  for (int i = 1; i < n; ++i) power[i] = power[i - 1] * p % m;
  for (int i = 0; i < n; ++i)
    h_s[i + 1] = (h_s[i] + (s[i] - 'a' + 1) * power[i]) % m;
  long long pre, suf;
  vector<pair<int, long long>> q;
  for (int i = 0; i < n; ++i) {
    pre = h_s[i + 1] % m;
    suf = (h_s[n] - h_s[n - i - 1] + m) % m;
    pre = pre * power[n - 1] % m;
    suf = suf * power[i] % m;
    if (pre == suf) q.push_back({i + 1, suf});
  }
  int beg = 0, end = q.size() - 1, mid, ans = -1;
  while (beg <= end) {
    mid = (beg + end) >> 1;
    auto it = poss(n, q[mid].second, q[mid].first, power, h_s);
    if (it) {
      beg = mid + 1;
      ans = q[mid].first;
    } else
      end = mid - 1;
  }
  if (ans == -1) {
    printf("Just a legend\n");
    return;
  }
  s[ans] = '\0';
  printf("%s\n", s);
  return;
}
int main() {
  scanf("%s", s);
  int n = strlen(s);
  solve(n);
  return 0;
}