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 size = 100000;
int sum[size + 10];
char str[size + 10];
stack<int> s;
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  while (cin >> (str + 1)) {
    int len = strlen(str + 1);
    int res = 0;
    int L = 0;
    int R = -1;
    for (int i = 1; i <= len; ++i) {
      sum[i] = sum[i - 1] + (str[i] == '[');
      if (s.empty()) {
        s.push(i);
      } else {
        if (str[s.top()] == '(' && str[i] == ')' ||
            str[s.top()] == '[' && str[i] == ']') {
          s.pop();
          if (s.empty()) {
            res = sum[i];
            L = 1, R = i;
          } else {
            if (res < sum[i] - sum[s.top()]) {
              res = sum[i] - sum[s.top()];
              L = s.top() + 1;
              R = i;
            }
          }
        } else {
          s.push(i);
        }
      }
    }
    cout << res << '\n';
    for (int i = L; i <= R; ++i) {
      cout << str[i];
    }
    cout << '\n';
    cout.flush();
  }
  return 0;
}
