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;
using ll = long long;
using ld = long double;
const int N = 1e6 + 10;
const ll INF = LLONG_MAX;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string second;
  cin >> second;
  int n = second.size(), ma = 0;
  vector<pair<char, int>> sul;
  for (int i = 0; i < n; i++) {
    if (second[i] == '[') ma++;
    if (sul.empty())
      sul.push_back(pair<char, int>(second[i], i));
    else {
      if (second[i] - sul.back().first == 2 ||
          second[i] - sul.back().first == 1)
        sul.pop_back();
      else
        sul.push_back(pair<char, int>(second[i], i));
    }
  }
  if (sul.empty()) {
    cout << ma << endl;
    if (ma) cout << second;
  } else {
    int temp = 0, j = 0, sum = 0;
    string q = "", w = "";
    for (int i = 0; i < n; i++) {
      if (i == sul[j].second && j < sul.size()) {
        if (temp > sum) {
          q = w;
          sum = temp;
        }
        temp = 0;
        w = "";
        j++;
      } else {
        w += second[i];
        if (second[i] == '[') temp++;
      }
    }
    if (temp > sum)
      cout << temp << endl << w;
    else
      cout << sum << endl << q;
  }
}
