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 = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. 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.

Input

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 105 characters.

Output

In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.

Examples

Input

([])


Output

1
([])


Input

(((


Output

0
#include <bits/stdc++.h>
using namespace std;
char s[123456];
int main() {
  scanf("%s", s);
  stack<pair<int, int> > st;
  int best = 0, rr = -1;
  int sqb = 0;
  for (int i = 0; s[i]; ++i) {
    if (s[i] == '(' || s[i] == '[') {
      st.push(make_pair(i, sqb));
      sqb = 0;
    } else if (s[i] == ')') {
      if (st.size() > 0 && s[st.top().first] == '(') {
        sqb += st.top().second;
        if (best < sqb) {
          best = sqb;
          rr = i;
        }
        st.pop();
      } else {
        stack<pair<int, int> >().swap(st);
        sqb = 0;
      }
    } else if (s[i] == ']') {
      if (st.size() > 0 && s[st.top().first] == '[') {
        sqb += st.top().second + 1;
        if (best < sqb) {
          best = sqb;
          rr = i;
        }
        st.pop();
      } else {
        stack<pair<int, int> >().swap(st);
        sqb = 0;
      }
    }
  }
  s[rr + 1] = 0;
  int ll = rr;
  stack<int> out;
  for (int i = rr; i >= 0; --i) {
    if (s[i] == ')' || s[i] == ']') {
      out.push(i);
    } else if (s[i] == '(') {
      if (out.size() > 0 && s[out.top()] == ')') {
        ll = i;
        out.pop();
      } else
        break;
    } else if (s[i] == '[') {
      if (out.size() > 0 && s[out.top()] == ']') {
        ll = i;
        out.pop();
      } else
        break;
    }
  }
  while (out.size() > 0) {
    ll = out.top() + 1;
    out.pop();
  }
  printf("%d\n", best);
  puts(s + ll);
  return 0;
}
