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 int N = 1e6 + 300;
const int INF = 1e9;
char S[N];
string s;
long long A = 416;
long long mod = 67280421310721;
long long mul(long long a, long long b) {
  long long res = (long double)a * b / mod;
  res = a * b - res * mod;
  while (res < 0) {
    res += mod;
  }
  while (res >= mod) {
    res -= mod;
  }
  return res;
}
long long pref[N];
long long pp[N];
int n;
vector<int> se;
bool check(int k) {
  long long v1 = (pref[n] - pref[n - k] + mod) % mod;
  for (int i = 1; i + k < n; i++) {
    long long vv = (pref[i + k] - pref[i] + mod) % mod;
    vv = mul(vv, pp[n - i - k]);
    if (vv == v1) {
      return 1;
    }
  }
  return 0;
}
int main() {
  scanf("%s", S);
  s = (string)S;
  n = s.length();
  srand(time(NULL));
  for (int i = 0; i < 0; i++) {
    if (rand() & 1) {
      mod = 1e18 + 3;
    } else {
      mod = 1e18 + 9;
    }
  }
  long long pow = 1ll;
  pp[0] = pow;
  for (int i = 0; i < n; i++) {
    pref[i + 1] = mul(pow, s[i] - 'a' + 1);
    pref[i + 1] = (pref[i + 1] + pref[i]) % mod;
    pow = mul(pow, A);
    pp[i + 1] = pow;
  }
  for (int i = 1; i <= n; i++) {
    long long v1 = (pref[n] - pref[n - i] + mod) % mod;
    long long v2 = mul(pref[i], pp[n - i]);
    if (v1 == v2) {
      se.push_back(i);
    }
  }
  int l = 0, r = se.size() - 1;
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if (check(se[mid])) {
      l = mid;
    } else {
      r = mid;
    }
  }
  for (int len = r; len >= l; len--) {
    if (check(se[len])) {
      printf("%s", s.substr(0, se[len]).c_str());
      return 0;
    }
  }
  printf("Just a legend");
}