Problem: 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

Solution (in PYTHON3):

Solution:
def longestPrefixSuffix(s):
    global d
    n = len(s)
    lps = [0] * n  # lps[0] is always 0

    # length of the previous
    # longest prefix suffix
    l = 0
    m = 0

    # the loop calculates lps[i]
    # for i = 1 to n-1
    i = 1
    while (i < n):
        if (s[i] == s[l]):
            l = l + 1
            lps[i] = l
            i = i + 1

        else:

            # (pat[i] != pat[len])
            # This is tricky. Consider
            # the example. AAACAAAA
            # and i = 7. The idea is
            # similar to search step.
            if (l != 0):
                l = lps[l - 1]

                # Also, note that we do
                # not increment i here

            else:

                # if (len == 0)
                lps[i] = 0
                i = i + 1

    s = s[::-1]
    lps1 = [0] * n  # lps[0] is always 0

    # length of the previous
    # longest prefix suffix
    l = 0

    # the loop calculates lps[i]
    # for i = 1 to n-1
    i = 1
    while (i < n):
        if (s[i] == s[l]):
            l = l + 1
            lps1[i] = l
            i = i + 1

        else:

            # (pat[i] != pat[len])
            # This is tricky. Consider
            # the example. AAACAAAA
            # and i = 7. The idea is
            # similar to search step.
            if (l != 0):
                l = lps1[l - 1]

                # Also, note that we do
                # not increment i here

            else:

                # if (len == 0)
                lps1[i] = 0
                i = i + 1
    ans=0
 
    for i in range(n-1):
        if lps[i]==lps1[n-(i-lps[i]+1)-1]:
            ans=max(ans,lps[i])

    return ans



# Driver program to test above function
q = input()

l=longestPrefixSuffix(q)
if l==0:
    print("Just a legend")
else:
    print(q[:l])