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>
struct Triple {
  int lt, rt, cnt;
};
Triple solve(std::string s) {
  const int n = (int)s.size();
  s = "?" + s;
  std::vector<int> stack, cnt(1 + n), leftPos(1 + n, n + 1);
  for (int i = 1; i <= n; ++i) {
    cnt[i] = cnt[i - 1] + (s[i] == '[');
  }
  int max = 0, lt = 1, rt = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '(' || s[i] == '[') {
      stack.push_back(i);
    } else if (stack.empty() || (s[stack.back()] == '(' && s[i] == ']') ||
               (s[stack.back()] == '[' && s[i] == ')')) {
      stack.clear();
    } else {
      int back = stack.back();
      stack.pop_back();
      int right = i;
      int left = leftPos[right] = std::min(back, leftPos[back - 1]);
      int curr = cnt[right] - cnt[left - 1];
      if (curr > max) {
        max = curr, lt = left, rt = right;
      }
    }
  }
  return Triple{lt - 1, rt - 1, max};
}
int main() {
  std::string s;
  while (std::cin >> s) {
    auto answ = solve(s);
    std::cout << answ.cnt << '\n'
              << s.substr(answ.lt, answ.rt - answ.lt + 1) << std::endl;
  }
  return 0;
}
