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;
const double pi = 3.141592653589793238460;
const long long mod = 1000000007;
const long long maxi = 200005;
const long long INF = INT_MAX;
const string no = "NO\n", yes = "YES\n";
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long x, unsigned long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
void mysol() {
  long long i = 0, j = 0, n, mx = 0, l = 0, r = 0;
  string s;
  cin >> s;
  n = s.length();
  vector<long long> Z(n, 0);
  for (i = 1; i < n; i++) {
    if (i <= r) Z[i] = min(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;
    }
  }
  set<long long> sol;
  for (i = 0; i < n; i++) {
    if (i + Z[i] == n) {
      sol.insert(Z[i]);
    }
  }
  for (i = 1; i < n; i++) {
    long long temp;
    if (i + Z[i] == n) {
      temp = Z[i] - 1;
    } else {
      temp = Z[i];
    }
    auto it = sol.upper_bound(temp);
    if (it == sol.begin()) {
      continue;
    } else {
      it--;
      temp = *it;
    }
    if (temp != 0 && temp > mx) {
      mx = temp;
      j = i;
    }
  }
  if (mx == 0)
    cout << "Just a legend\n";
  else
    cout << s.substr(j, mx);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  while (t--) {
    mysol();
  }
}