A bracket sequence is a string, containing only characters "(", ")", "[" and "]".

A correct bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()[]", "([])" are correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.

A substring *s*[*l*... *r*] (1 ≤ *l* ≤ *r* ≤ |*s*|) of string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (where |*s*| is the length of string *s*) is the string *s*<sub class="lower-index">*l*</sub>*s*<sub class="lower-index">*l* + 1</sub>... *s*<sub class="lower-index">*r*</sub>. The empty string is a substring of any string by definition.

You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 10<sup class="upper-index">5</sup> characters.
s = input()
a = []
for i in range(len(s)):
    if len(a) != 0 and ((s[a[-1]] == '(' and s[i] == ')') or (s[a[-1]] == '[' and s[i] == ']')):
        a.pop()
    else:
        a.append(i)
if len(a) == 0:
    print(s.count('['))
    print(s)
else:
    s1 = s[0: a[0]]
    le = s[0: a[0]].count('[')
    for i in range(len(a) - 1):
        le1 = s[a[i] + 1: a[i + 1]].count('[')
        if le1 > le:
            s1 = s[a[i] + 1: a[i + 1]]
            le = le1
    le1 = s[a[-1] + 1:].count('[')
    if le1 > le:
        s1 = s[a[-1] + 1:]
        le = le1
    print(le)
    print(s1)