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;
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
int main() {
  string s;
  stack<pair<char, int> > a;
  cin >> s;
  vector<pair<int, int> > vec;
  vector<int> rectangle;
  int numrect = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(') {
      a.push({'(', i});
    }
    if (s[i] == '[') {
      a.push({'[', i});
    }
    if (s[i] == ')') {
      if (!a.empty() && a.top().first == '(') {
        pair<int, int> k = a.top();
        a.pop();
        vec.push_back({k.second, i});
      } else {
        while (!a.empty()) a.pop();
      }
    }
    if (s[i] == ']') {
      if (!a.empty() && a.top().first == '[') {
        pair<int, int> k = a.top();
        a.pop();
        vec.push_back({k.second, i});
        numrect++;
      } else {
        while (!a.empty()) a.pop();
      }
    }
    rectangle.push_back(numrect);
  }
  int l = 0, r = 0;
  sort(vec.begin(), vec.end());
  int n = vec.size();
  int flag = 0;
  vector<pair<int, int> > vec1;
  for (int i = 0; i < n; i++) {
    l = vec[i].first;
    r = vec[i].second;
    vec1.push_back({l, r});
    if (i == n - 1) break;
    while (vec[i + 1].first > l && vec[i + 1].second < r) {
      i++;
    }
  }
  sort(vec1.begin(), vec1.end(), sortbysec);
  cout << endl;
  n = vec1.size();
  for (int i = 0; i < n - 1; i++) {
    l = vec1[i].first;
    r = vec1[i].second;
    while (vec1[i].second + 1 == vec1[i + 1].first) {
      r = vec1[i + 1].second;
      i++;
    }
    if (l != vec1[i].first) vec1.push_back({l, r});
  }
  int p = 0, q = 0;
  int maxrect = 0;
  for (auto x : vec1) {
    if (rectangle[x.second] - rectangle[x.first] >= maxrect) {
      maxrect = rectangle[x.second] - rectangle[x.first];
      p = x.first;
      q = x.second;
    }
  }
  if (vec.size() == 0) {
    cout << 0;
    return 0;
  }
  cout << maxrect << endl;
  for (int i = p; i <= q; i++) {
    cout << s[i];
  }
}
