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.Arrays;
import java.util.Scanner;

public class Main {
	static int len;
	static String s;
	static int[] next;
	static char[] c;
	static boolean[] vis;

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		s = in.next();
		len = s.length();
		c = s.toCharArray();
		vis = new boolean[len + 1];
		next = new int[len + 1];
		getNext(c, next);
		int i = len, m = 0, index = 0;
		while (next[i] != 0) {
			vis[next[i]] = true;
			i = next[i] ;
		}
		index = 0;
		for (int j = 1; j < len; j++) {
			if (vis[next[j]] && next[j] > m) {
				m = next[j];
				index = 1;
			}
		}
		if (index == 1)
			System.out.println(s.substring(0, m));
		else
			System.out.println("Just a legend");
	}


	static void getNext(char[] p, int[] f) {
		int m = p.length;
		f[0] = f[1] = 0;
		for (int i = 1; i < m; ++i) {
			int j = f[i];
			while (j > 0 && p[i] != p[j])
				j = f[j];
			f[i + 1] = p[i] == p[j] ? 1 + j : 0;
		}
	}

	public static void newnext(char[] c) {
		int j = -1, i = 0;
		next[0] = -1;
		while (i < len) {
			if (j == -1 || c[i] == c[j]) {
				i++;
				j++;
				next[i] = j;
			} else {
				j = next[j];
			}
		}

	}// 优化KMPnext算法

}