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
import java.util.Arrays;
import java.util.Scanner;

public class task1 {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        String str = scan.nextLine();
        int[] s = Fix(str);
        int lfix = 0;
        for(int i=1;i<str.length();i++) {
                if(s[i] == str.length()-i && lfix >= s[i]) {
                        System.out.println(str.substring(0, s[i]));
                        return;
                }
                lfix = Math.max(lfix, s[i]);
        }
        System.out.println("Just a legend");
    }
    public static int[] Fix(String string) {
        char[] a = string.toCharArray();
        int[] s = new int[string.length()];
        int n = string.length();
        int beg = 0, fin = 0;
        s[0] = n;
        for (int i=1;i<n;i++) {
            if (i>fin) {
                beg=i;
                fin=i-1;
                while(fin+1<n && a[fin+1] == a[fin+1-beg]){
                    fin++;
                }
                s[i]=fin-beg+1;
            } else{
                int k=i-beg;
                if (s[k]<fin-i+1) {
                    s[i]=s[k];
                } 
                else{
                    beg=i;
                    while(fin+1<n && a[fin+1] == a[fin +1-beg]) {
                        fin++;
                    }
                    s[i]=fin-beg+1;
                }
            }
        }
        return s;
    }
}