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
/* Adapted from the solution of sevenkplus
   http://www.codeforces.com/contest/126/submission/845264
   Very compact implementation of the rabin-karp method
   (originally in C++)

   Commented and translated to Java by Danny Sleator  (aka Darooha)
*/
import java.io.*;
import java.util.*;

public class D_rabin_karp {
    static final long P = 1000000007;
    static long[] p;  // p[i] = P^i modulo 2^64
    static long[] a;  // a[i] = s[i-1]*p[0] + s[i-2]*p[1] + ... + s[0]*p[i-1]
    static int n;


    static long hh(int x, int y) {
	/*  Assumes x<=y.  Let k = y-x.  This function returns
	    s[x]*p[0] + s[x+1]*p[1] + ... + s[y]*p[k].
	    In other words, it's the hash function from x to y inclusive
	 */
        return a[y+1]-a[x]*p[y-x+1];
    }

    static int ff (int x) {
	/* This returns the longest string length l, starting
	   at position x, such that s[x -- x+l-1] = s[0 -- l-1].
	   Consider lengths that don't include the last character.
	*/
        int L=0, R=n-x-1;
        while(L<R) {
	    int M=(L+R+1)/2;
	    if (hh(0,M-1)==hh(x,x+M-1)) L=M; else R=M-1;
	}
        return R;
    }

    public static void main(String[] args) throws IOException {
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	String input = in.readLine();
	char[] s= input.toCharArray();
	n = s.length;
	p = new long[n+1];
	a = new long[n+1];
	int S = 0;

        p[0]=1;
	for(int i=1; i<=n; i++) p[i] = p[i-1] * P;
        for(int i=1; i<=n; i++) a[i]=a[i-1]*P+s[i-1];
        for(int i=1; i<n-1; i++) S = Math.max(S, ff(i));
        for(int i=S; i>=1; i--) 
	    if (hh(0,i-1) == hh(n-i,n-1)) {
		System.out.println(input.substring(0, i));
		System.exit(0);
	    }
        System.out.println("Just a legend");
    }
}
