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;
const int maxn = 100015;
int last[maxn], num[maxn], g[200];
stack<pair<char, int> > stk;
int main() {
  int i, j, n, mn = -1, ss = 0, f = -1;
  cin >> s;
  n = s.length();
  g['['] = ']';
  g[']'] = '[';
  g['('] = ')';
  g[')'] = '(';
  for (i = 0; i < n; i++) last[i] = i + 1;
  for (i = 0; i < n; i++) {
    if (s[i] == '[' || s[i] == '(') {
      stk.push(make_pair(s[i], i));
    } else {
      if (stk.empty() || stk.top().first != g[s[i]]) {
        while (!stk.empty()) stk.pop();
      } else {
        last[i] = stk.top().second;
        stk.pop();
      }
    }
  }
  for (i = 1; i < n; i++) {
    if (last[i]) last[i] = min(last[i], last[last[i] - 1]);
  }
  for (i = 1; i <= n; i++) num[i] = num[i - 1] + (s[i - 1] == '[');
  for (i = 0; i < n; i++) {
    if (num[i + 1] - num[last[i]] > mn) {
      mn = num[i + 1] - num[last[i]];
      ss = last[i];
      f = i;
    }
  }
  cout << mn << endl;
  for (i = ss; i <= f; i++) putchar(s[i]);
}
