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>
int n, m;
int next[100010], stiv[100010], nr[100010];
char s[100010];
int main() {
  scanf("%s", s + 1);
  s[0] = ' ';
  n = strlen(s) - 1;
  for (int i = 1; i <= n; i++) {
    stiv[++m] = i;
    if (m > 1 && s[stiv[m]] == ')' && s[stiv[m - 1]] == '(') {
      next[stiv[m - 1]] = i;
      m -= 2;
    }
    if (m > 1 && s[stiv[m]] == ']' && s[stiv[m - 1]] == '[') {
      next[stiv[m - 1]] = i;
      m -= 2;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (s[i] == '[') nr[i]++;
    nr[i] += nr[i - 1];
  }
  int pmax = -1;
  for (int i = n; i > 0; i--)
    if (next[i]) {
      if (next[next[i] + 1] != 0) next[i] = next[next[i] + 1];
      if (pmax == -1 || nr[next[i]] - nr[i - 1] > nr[next[pmax]] - nr[pmax - 1])
        pmax = i;
    }
  if (pmax < 0)
    printf("0\n");
  else {
    printf("%d\n", nr[next[pmax]] - nr[pmax - 1]);
    for (int i = pmax; i <= next[pmax]; i++) printf("%c", s[i]);
    printf("\n");
  }
  return 0;
}
