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 s[100005];
int match[100005], sum[100005];
bool isMatch(char a, char b) {
  if (a == '(' && b == ')') return true;
  if (a == '[' && b == ']') return true;
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  scanf("%s", s);
  int n = strlen(s);
  int best = -1, best_i = 0, best_j = 0;
  sum[n] = 0;
  for (int i = n - 1; i >= 0; --i) {
    match[i] = i;
    if (i + 1 < n && isMatch(s[i], s[match[i + 1]]))
      match[i] = match[i + 1] + 1;
    if (i < match[i] && match[i] < n) match[i] = match[match[i]];
    sum[i] = sum[i + 1] + (s[i] == '[');
    if (sum[i] - sum[match[i]] > best) {
      best = sum[i] - sum[match[i]];
      best_i = i;
      best_j = match[i];
    }
  }
  s[best_j] = '\0';
  printf("%d\n%s\n", best, s + best_i);
}
