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;
char ln[100009];
int sm[100009];
int main() {
  gets(ln + 1);
  int i, j, k, t = 0;
  stack<pair<int, int> > st;
  st.push(pair<int, int>(-1, 0));
  for (i = 1; ln[i]; i++) {
    sm[i] = sm[i - 1];
    if (ln[i] == '(') {
      st.push(pair<int, int>(1, i));
    } else if (ln[i] == '[') {
      st.push(pair<int, int>(2, i));
      sm[i]++;
    } else if (st.size() && ((st.top().first == 1 && ln[i] == ')') ||
                             ((st.top().first == 2 && ln[i] == ']')))) {
      st.pop();
      if (sm[i] - sm[st.top().second] > t) {
        t = sm[i] - sm[st.top().second];
        j = st.top().second + 1;
        k = i;
      }
    } else {
      st.push(pair<int, int>(-1, i));
    }
  }
  cout << t << '\n';
  if (t) {
    ln[k + 1] = 0;
    puts(ln + j);
  }
  return 0;
}
