Algorithm

B. Password
time limit per test
2 seconds
memory limit per test
256 megabytes
Problem Link- https://codeforces.com/problemset/problem/126/B
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
Copy
fixprefixsuffix
output
Copy
fix
input
Copy
abcdabc
output
Copy
Just a legend



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <iostream>
#include <string>

int main(){

    std::string s; getline(std::cin, s);
    std::string ans("");
    for(int p = 1; p  < s.size(); p++){
        std::string prefix = s.substr(0, p);
        std::string suffix = s.substr(s.size() - p, p);
        if(prefix != suffix){continue;}

        int midPos = s.find(prefix, 1);
        if(midPos == std::string::npos || midPos >= (s.size() - p)){continue;}

        ans = prefix;
    }

    std::cout  < < ((ans.size() > 0) ? ans : "Just a legend")  < < std::endl;

    return 0;
}
Copy The Code & Try With Live Editor

Input

x
–
+
cmd
fixprefixsuffix

Output

x
–
+
cmd
fix
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Demonstration

Codeforces Solution-B. Password-Solution in C, C++, Java, Python

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