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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class CF126BPassword {

	public static void main(String[]args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String s=br.readLine();
		findPassword(s);
		
	}
	public static void findPassword(String s){
		int [] prefixArray=computePrefixArray(s);
		//System.out.println(Arrays.toString(prefixArray));
		//check the last if zero means no soloution 
		//if not search for the sane pattrer ni the array
		//if not found ,search for the sec best pattern which is the index of prefixarrya -1
		//if not sero then its the soloution
		int last=prefixArray[s.length()-1];
		boolean found=false;
		if(last==0)
			System.out.println("Just a legend");
		else {
			for(int i=s.length()-2;i>0;i--)
				if(prefixArray[i]==last)
					{found=true;
					 System.out.println(s.substring(s.length()-last));
					break;
					}
			if(!found)
				if(prefixArray[last-1]!=0)
					System.out.println(s.substring(s.length()-prefixArray[last-1]));
					
				else System.out.println("Just a legend");
		}
			
		
		
		
	}
	public static int[] computePrefixArray(String pattern){
		int [] prefixArray=new int[pattern.length()];
		prefixArray[0]=0;
		int j=0;
		for(int i=1;i<prefixArray.length;){
			if(pattern.charAt(i)==pattern.charAt(j))
				{prefixArray[i]=j+1;
				 j++;
				 i++;
				}
			else {
				if(j!=0)
					j=prefixArray[j-1];
				else{
					prefixArray[i]=0;
				    i++;
				}
				
				
			}
			
			
			
		}
		return prefixArray;
		
	}
}
