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;
string s;
int st[100100], ns = 1;
int num[100100];
int rst[100100];
int res = 0, l = 0, r = -1;
int main() {
  cin >> s;
  st[0] = -5;
  int i;
  num[0] = -1;
  int cur = 0;
  for (i = 0; i < (int)s.length(); i++) {
    if (s[i] == '[') {
      cur++;
      st[ns] = 1;
      num[ns] = i;
      rst[ns] = cur;
      ns++;
    }
    if (s[i] == '(') {
      st[ns] = 2;
      num[ns] = i;
      rst[ns] = cur;
      ns++;
    }
    if (s[i] == ']' || s[i] == ')') {
      int g;
      if (s[i] == ']')
        g = -1;
      else
        g = -2;
      if (st[ns - 1] + g == 0) {
        ns--;
        if (cur - rst[ns - 1] > res) {
          res = cur - rst[ns - 1];
          l = num[ns - 1] + 1;
          r = i;
        }
      } else {
        ns = 1;
        num[0] = i;
        cur = 0;
      }
    }
  }
  cout << res << endl;
  for (i = l; i <= r; i++) cout << s[i];
}
