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;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  string str;
  cin >> str;
  long long sz = str.size();
  long long a[sz + 1], cnt[sz + 1];
  cnt[0] = 0;
  for (long long i = 0; i < sz; i++) {
    cnt[i + 1] = cnt[i];
    if (str[i] == '(')
      a[i + 1] = 1;
    else if (str[i] == ')')
      a[i + 1] = 2;
    else if (str[i] == '[')
      a[i + 1] = 3;
    else
      a[i + 1] = 4, cnt[i + 1]++;
  }
  stack<pair<long long, long long> > s;
  long long op[sz + 1];
  memset(op, -1, sizeof(op));
  for (long long i = 1; i <= sz; i++) {
    if (a[i] == 1 || a[i] == 3) {
      s.push({a[i], i});
    } else if (s.empty()) {
      continue;
    } else if ((a[i] == 2 && s.top().first == 1) ||
               (a[i] == 4 && s.top().first == 3)) {
      op[i] = s.top().second;
      s.pop();
    } else {
      while (!s.empty()) s.pop();
    }
  }
  long long mx = 0, start = -1, end = -1;
  for (long long i = 1; i <= sz; i++) {
    if (op[i] == -1) continue;
    if (op[op[i] - 1] != -1) op[i] = op[op[i] - 1];
    if (cnt[i] - cnt[op[i] - 1] >= mx) {
      start = op[i];
      end = i;
      mx = cnt[i] - cnt[op[i] - 1];
    }
  }
  cout << mx << "\n";
  if (start != -1) {
    for (long long i = start; i <= end; i++) cout << str[i - 1];
  }
}
