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 nmax = 100002;
int best[nmax];
int start[nmax];
char rev[256];
int main() {
  int i, j, k, n, m;
  string s;
  cin >> s;
  n = s.length();
  rev[')'] = '(';
  rev[']'] = '[';
  int ansi = 0, anslen = 0, ans = 0;
  stack<int> st;
  for (i = 0; i < n; i++) {
    if (s[i] == '(' || s[i] == '[') st.push(i);
    if (s[i] == ')' || s[i] == ']')
      if (st.empty() || s[st.top()] != rev[s[i]]) {
        while (!st.empty()) st.pop();
      } else {
        k = st.top();
        st.pop();
        best[i] = (s[i] == ']') + best[i - 1] + best[k - 1];
        start[i] = (best[k - 1] == 0 ? k : start[k - 1]);
        if (ans < best[i]) {
          ans = best[i];
          ansi = start[i];
          anslen = i - start[i] + 1;
        }
      }
  }
  cout << ans << endl;
  cout << s.substr(ansi, anslen) << endl;
  return 0;
}
