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
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF127D extends PrintWriter {
	CF127D() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF127D o = new CF127D(); o.main(); o.flush();
	}

	void main() {
		byte[] cc = sc.next().getBytes();
		int n = cc.length;
		int[] ff = new int[n];
		for (int i = 0, j = 1; j < n; i++, j++) {
			while (cc[i] != cc[j]) {
				if (i == 0) {
					i = -1;
					break;
				}
				i = ff[i - 1];
			}
			ff[j] = i + 1;
		}
		int f = ff[n - 1];
		if (f > 0) {
			boolean dup = false;
			for (int i = 1; i < n - 1; i++)
				if (ff[i] >= f) {
					dup = true;
					break;
				}
			if (!dup)
				f = ff[f - 1];
		}
		println(f == 0 ? "Just a legend" : new String(cc, 0, f));
	}
}
