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.
a = input()
stek = []
l = 0
r = 0
count_scob = 0
max_count = 0
last = 0
for i in range(len(a)):
    if a[i] == '[' or a[i] == '(':
        stek.append(i)
        last = a[i]
    elif (a[i] == ')' and last == '(') or (a[i] == ']' and last == '['):
        count_scob = a[stek[-1]:i + 1].count('[')
        if count_scob >= max_count:
            l = stek[-1]
            r = i
            stek.pop()
            if stek != []:
                last = a[stek[-1]]
            max_count = count_scob
if max_count == 0:
    print(0)
else:
    print(max_count)
    print(a[l:r + 1])
