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.
x=input()
ans=[]
s=['a']
tmp=''
count1=0
count2=0
co=0
for i in x:
    if i==')':
        if s[-1]=='(':
            del s[-1]
        else:
            if len(s)==2:
                ans.append(tmp[:-1])
                tmp=''
            else:
                s=['a']
                tmp=''
    elif i==']':
        if s[-1]=='[':
            del s[-1]
        else:
            if len(s)==2:
                ans.append(tmp[:-1])
                tmp=''
            else:
                s=['a']
                tmp=''
    else:
        s.append(i)
    tmp+=i
    co+=1
    if len(s)==1 and co==len(x):
        ans.append(tmp)
q=0
w=1
if len(ans)>0:
    while w<len(ans):
        if ans[q].count('[')>=ans[w].count('['):
            w+=1
        else:
            q=w
            w+=1
    print(ans[q].count('['))
    print(ans[q])
else:
    print('0')

    
    
