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 str[100010];
stack<int> sta;
int sum[100010];
int main() {
  cin >> str;
  int len = strlen(str);
  int i, j;
  i = 0;
  while (i < len) {
    if (str[i] == '(' || str[i] == '[') {
      sta.push(i);
      break;
    } else
      i++;
  }
  if (i == len) {
    cout << 0 << endl;
    return 0;
  }
  int ans = 0;
  int x, y;
  for (i++; i < len; i++) {
    sum[i] = sum[i - 1] + (str[i] == ']');
    if (!sta.empty()) {
      char ch = str[sta.top()];
      if ((ch == '(' && str[i] == ')') || (ch == '[' && str[i] == ']')) {
        sta.pop();
        if (sta.empty() && sum[i] > ans) {
          ans = sum[i];
          x = 0;
          y = i;
        } else if (!sta.empty() && ans < sum[i] - sum[sta.top()]) {
          ans = sum[i] - sum[sta.top()];
          x = sta.top() + 1;
          y = i;
        }
      } else
        sta.push(i);
    } else
      sta.push(i);
  }
  cout << ans << endl;
  if (ans) {
    for (i = x; i <= y; i++) {
      cout << str[i];
    }
    cout << endl;
  }
  return 0;
}
