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.
s, st, v, vi, vj, vc = input(), [], [], 0, 0, 0
for i, ch in enumerate(s):
    if ch in '[(':
        st.append(i)
    else:
        if st and s[st[-1]] + ch in ('()', '[]'):
            b = (st[-1], i + 1)
            if v and v[-1][1] == i:
                v[-1] = b
            else:
                v.append(b)
            if len(v) >= 2 and v[-2][1] == v[-1][0]:
                v[-2:] = [(v[-2][0], v[-1][1])]
            st.pop()
        else:
            st = []
for b in v:
    c = s.count('[', b[0], b[1])
    if c > vc:
        vi, vj, vc = b[0], b[1], c
print(vc)
print(s[vi:vj])
