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.
class Stack:
    def __init__(self,sz):
        self.top=-1
        self.st=[None for x in range(0,sz)]
    def push(self,value):
        self.top+=1
        self.st[self.top]=value
        
    def peek(self):
        return self.st[self.top]

    def pop(self): 
        last=self.st[self.top]
        self.top-=1
        return last
    def is_empty(self):
        return self.top==-1

def balance(stk,size):
    stack=Stack(size)
    for i in stk:
        if i=="{" or i=="(" or i=="[" :
            stack.push(i)
        else:
            if stack.is_empty():
                return "NO"
            if i=="}" and stack.peek()!="{":
                return "NO"
            if i==")" and stack.peek()!="(":
                return "NO"
            if i=="]" and stack.peek()!="[":
                return "NO"
            else:
                stack.pop()
    if stack.is_empty():
        return "YES"
    
    return "NO"
            

stk=input()
size=len(stk)
val=balance(stk,size)
if val=="YES":
    print(1)
    print(stk)
else:
    print(0)
