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;
const int oo = 1e9;
char match(char c) { return c == ']' ? '[' : '('; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string line;
  while (cin >> line) {
    int n = line.size();
    vector<int> closing(n, -1);
    stack<int> st;
    pair<int, string> ans = pair<int, string>(0, "");
    for (int i = (0); i < ((int)line.size()); ++i) {
      char c = line[i];
      if (c == '(' || c == '[') {
        st.push(i);
      } else {
        if (!st.empty() && line[st.top()] == match(c)) {
          closing[st.top()] = i;
          st.pop();
        } else {
          while (!st.empty()) st.pop();
        }
      }
    }
    int l = 0, r = 0;
    while (l < n) {
      while (r < n && closing[r] == -1) r++;
      l = r;
      while (r < n && closing[r] != -1) r = closing[r] + 1;
      string s = line.substr(l, r - l);
      int cnt = count(s.begin(), s.end(), '[');
      if (cnt > ans.first) {
        ans.first = cnt;
        ans.second = s;
      }
      l = r;
    }
    cout << ans.first << '\n';
    cout << ans.second << '\n';
  }
  return 0;
}
