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;
stack<pair<char, int> > S;
string g;
int t[100100];
int main() {
  cin >> g;
  t[0] = (g[0] == '[');
  for (int i = 1; i < g.size(); i++) t[i] = t[i - 1] + (g[i] == '[');
  int from = -1, to = -1, mmax = 0, ffrom = 0, tto;
  for (int i = 0; i < g.size(); i++) {
    if (g[i] == '[' || g[i] == '(') {
      S.push(make_pair(g[i], i));
    } else {
      if (S.empty()) {
        ffrom = i + 1;
        continue;
      }
      if ((g[i] == ']' && S.top().first == '[') ||
          ((g[i] == ')' && S.top().first == '('))) {
        S.pop();
        int tefrom;
        if (S.size() == 0)
          tefrom = ffrom;
        else
          tefrom = S.top().second + 1;
        int now = t[i] - (tefrom == 0 ? 0 : t[tefrom - 1]);
        if (now >= mmax) {
          mmax = now;
          from = tefrom;
          to = i;
        }
      } else {
        while (!S.empty()) S.pop();
        ffrom = i + 1;
      }
    }
  }
  printf("%d\n", mmax);
  if (from != -1)
    for (int i = from; i <= to; i++) printf("%c", g[i]);
  puts("");
  return 0;
}
