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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import static java.lang.System.in;
import java.math.BigInteger;
import java.util.PriorityQueue;
import java.util.Stack;
import java.util.stream.Stream;
 
 
public class Main {
    
    
    
    public static void main(String[] args) throws IOException, Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        bw.append(solve(br.readLine())+"\n");
        bw.close();
        
    }

    private static String solve(String S) {
        int P = 31;
        int mod = (int) (1e9+7);
        int[] pow = new int[(int)(1e6+1)];
        //compute the powers
        pow[0]=1;
        for(int i=1;i<pow.length;i++) pow[i]=(int) ((1l*pow[i-1]*P)%mod);
        //System.out.println("done calculating powers");
        //compute the hash prefixes
        int n = S.length();
        int[] hashes = new int[S.length()];
        for(int i=0;i<S.length();i++) {
            hashes[i] = (int) ((1l*(S.charAt(i)-'a'+1)*pow[i])%mod);
            if(i>0) hashes[i]+=hashes[i-1];
            hashes[i]%=mod;
        }
       // System.out.println("done calculatiing hash prefixes");
        //iterate over the length and solve the problem
        boolean seen=false;
        String res = "Just a legend";
        for(int L=n-2; L>=1; L--) {
           // System.out.println("strings "+S.substring(0, L)+" and "+S.substring(n-L));
            //compute the hash for both the prefix and the suffix
            int hp = (int) ((1l*hashes[L-1]*pow[n-L])%mod);
            int hs = (int) ((1l*hashes[n-1]-1l*hashes[n-L-1]+mod)%mod);
            
            if(hp==hs) {
                
                //calculate the other hashes for substrings at the middle
                for(int j=1;j<n-L;j++) {
                    //j till j+L-1
                    int hj = (int) ((1l*hashes[j+L-1]-1l*hashes[j-1]+mod)%mod);
                    
                    if(hj == (int) ((1l*hashes[L-1]*pow[j])%mod) ) {
                        seen=true;
                        res = S.substring(0, L);
                        break;
                    }
                }
            }
            
            if(seen) break;
        }
        return res;
    }
    
}