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[111111];
int ans[111111] = {0}, f[111111] = {0}, res = 0, resl, resr, len;
int main() {
  scanf("%s", str);
  len = strlen(str);
  for (int i = 0; i <= len; i++) {
    if (i == 0 || i - f[i - 1] - 1 < 0) continue;
    if (str[i] == ')' && str[i - f[i - 1] - 1] == '(') {
      f[i] = f[i - 1] + 2 + f[i - f[i - 1] - 2];
      ans[i] = ans[i - 1] + ans[i - f[i - 1] - 2];
    }
    if (str[i] == ']' && str[i - f[i - 1] - 1] == '[') {
      f[i] = f[i - 1] + 2 + f[i - f[i - 1] - 2];
      ans[i] = ans[i - 1] + ans[i - f[i - 1] - 2] + 1;
    }
    if (ans[i] > res) res = ans[i], resr = i, resl = i - f[i] + 1;
  }
  printf("%d\n", res);
  if (res == 0) return 0;
  for (int i = resl; i <= resr; i++) printf("%c", str[i]);
  printf("\n");
  return 0;
}
