B. Password
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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

题意:给出最长的前缀使得:

1.前缀和后缀相同

2.出现次数大于3


思路:  出现次数用cnt[next[i]]+=cnt[i]解决

前缀后缀相同 :  利用next[next[i]]是次长公共前缀解决


AC代码:

#include <stdio.h>
#include <iostream>
#include <string.h>
typedef long long ll;

using namespace std;

const int N=1e6+6;
int _next[N];
int lenb;
char b[N];
int cnt[N];

void set_naxt()
{
    int i=0,j=-1;
    _next[0]=-1;
    while(i<lenb)
    {
        if(j==-1||b[i]==b[j])
        {
            i++; j++;
            _next[i]=j;
        }
        else
        j=_next[j];
    }
}

int main(void){
    scanf("%s",b);
    lenb=strlen(b);
    set_naxt();
    //for(int i=0;i<=lenb;i++) printf("%d ",_next[i]);
    for(int i=1;i<=lenb;i++)    cnt[i]=1;
    for(int i=lenb;i>=1;i--) cnt[_next[i]]+=cnt[i];
    //for(int i=1;i<=lenb;i++)    printf("cnt[%d]=%d\n",i,cnt[i]);
    int len=lenb;
    while(len){
        if(cnt[len]>=3){
            for(int i=0;i<len;i++)  printf("%c",b[i]);
            return 0;
        }
        len=_next[len];
    }
    cout <<"Just a legend"<<endl;
}