Password CodeForces - 126B [KMP+DP思维]

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 
#include 
#include 
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=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