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 MAX_N = 2e5 + 7;
int x, cnt[MAX_N], fc[MAX_N], l[MAX_N];
char seq[MAX_N];
int main() {
  cin >> seq + 1;
  vector<pair<char, int> > stack;
  int ans = 0, x = 0, y = 0;
  for (int i = 1; seq[i]; i++) fc[i] = fc[i - 1] + (seq[i] == '[');
  for (int i = 1; seq[i]; i++) {
    if (seq[i] == '[' || seq[i] == '(')
      stack.push_back(make_pair(seq[i], i));
    else if (stack.size()) {
      char c = stack.back().first;
      int s = stack.back().second;
      if ((seq[i] == ']' && c != '[') || (seq[i] == ')' && c != '('))
        stack.clear();
      else {
        stack.pop_back(), cnt[i] = cnt[s - 1] + fc[i] - fc[s - 1];
        l[i] = (l[s - 1] == 0) ? s : l[s - 1];
        if (cnt[i] > ans) x = l[i], y = i, ans = cnt[i];
      }
    }
  }
  cout << ans << endl;
  if (ans)
    for (int i = x; i <= y; i++) cout << seq[i];
  else
    cout << "";
  cout << endl;
  return 0;
}
