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.

# coding: utf-8

# In[12]:

def op(x):
    if x == '[' or x == '(':
        return True
    return False

def clos(x,y):
    if x == '[' and y == ']':
        return True
    if x == '(' and y == ')':
        return True
    return False


# In[13]:

def solver(seq):
    stack = []
    max_brack = 0
    brack = 0
    
    for el in seq:
        if stack:
            if op(el):
                stack.append(el)
            else:
                if clos(stack[-1],el):
                    if el == ']':
                        brack += 1
                else:
                    if brack > max_brack:
                        max_brack = brack
                    brack = 0
                    stack = []
        else:
            if op(el):
                stack.append(el)
    return max_brack

# In[ ]:

a = input()
print(solver(a))

