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
def prefix(s):
    p = [0]
    for i in s[1:]:
        j = p[-1]
        while j > 0 and s[j] != i:
            j = p[j - 1]
        p.append(j + (s[j] == i))
    return p


def kmp(s, p):
    s = p + '$' + s
    pre = prefix(s)
    o = []

    for i in range(len(p), len(pre)):
        if pre[i] == len(p):
            o.append(i - 2 * len(p))

    return o


s = input()
p = prefix(s)

print(s[:p[-1]] if p[-1] and p.count(p[-1]) > 1 and len(s) >
      2 else s[:p[p[-1] - 1]] if p[p[-1] - 1] else 'Just a legend')
