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.
import sys
inpu = sys.stdin.readline
prin = sys.stdout.write
s = inpu().rstrip('\n')
stack = []
brackets = {'(', '['}
check = [0]*(len(s))
ind = []
for j, i in enumerate(s) :
    if i in brackets :
        stack.append(i)
        ind.append(j)
    elif len(stack) == 0:
        continue
    elif i == ']' :
        if stack[-1] == '[' :
            check[ind.pop()] = 1
            check[j] = 1
            stack.pop()
        else :
            stack = []
            ind = []
    else :
        if stack[-1] == '(' :
            check[ind.pop()] = 1
            check[j] = 1
            stack.pop()
        else :
            stack = []
            ind = []
ans = [0]*len(check)
for i in range(len(check)) :
    if check[i] != 0 :
        ans[i] = ans[i - 1] + 1
ind = 0
ma = 0
for i in range(len(ans)) :
    if ans[i] > ma :
        ma = ans[i]
        ind = i
ans = s[ind - ma + 1:ind + 1]
prin(str(ans.count('[')) + '\n' + ans + '\n')