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;
using ll = long long;
const ll inf = 2e10;
ll bigmod(ll a, ll b) {
  if (b == 0) return 1;
  ll x = bigmod(a, b / 2);
  x = x * x;
  if (b & 1) x = a * x;
  return x;
}
vector<ll> z;
void Z(string s) {
  ll n = s.length();
  z.resize(n);
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min((ll)r - i + 1, z[i - l]);
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return;
}
void solve() {
  string s;
  ll cnt = 0, mx = 0;
  cin >> s;
  bool ok = 1;
  if (s.size() < 3)
    ok = 0;
  else {
    Z(s);
    for (ll i = 0; i < s.size(); i++) {
      if (z[i] + i == s.size() and mx >= z[i]) {
        cnt = z[i];
        break;
      }
      mx = max(mx, z[i]);
    }
    if (cnt == 0) ok = 0;
  }
  if (ok) {
    for (ll i = 0; i < cnt; i++) cout << s[i];
  } else
    cout << "Just a legend";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << '\n';
  return 0;
}