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.util.*;
import java.io.*;

public class File {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        // Substring t, that is not the prefix or suffix,
        // that is also a prefix and suffix of the string.
        String s = sc.next();
        
        int[] lps = KMP(s);

        int N = s.length();
        if (lps[N-1] == 0) {
            System.out.println("Just a legend");
            return;
        }

        // Check for substring of length lps[N-1].
        int suffixLength = lps[N-1];
        for (int i = 1; i + 1 < s.length(); i++) {
            if (lps[i] == suffixLength) {
                System.out.println(s.substring(0, suffixLength));
                return;
            }
        }

        // None, then check for substring of length lps[lps[N-1]]
        suffixLength = lps[lps[N-1]-1];
        if (suffixLength == 0) {
            System.out.println("Just a legend");
            return;
        }
        
        for (int i = 1; i + 1 < s.length(); i++) {
            if (lps[i] == suffixLength) {
                System.out.println(s.substring(0, suffixLength));
                return;
            }
        }

        System.out.println("Just a legend");    
    }

    // Modify to exclude overlapping prefix/suffixes.
    public static int[] KMP(String s) {
        int N = s.length();
        int[] lps = new int[N];

        // lps stores length of longest prefix
        // this substring equals to.
        int pIndex = 0;
        int i = 1;

        while (i < N) {
            if (s.charAt(i) == s.charAt(pIndex)) {
                pIndex++;
                lps[i] = pIndex;
                i++;
            }
            else {
                if (pIndex == 0) {
                    i++;
                }
                else {
                    pIndex = lps[pIndex-1];
                }
            }
        }

        return lps;
    }
}