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 double eps = 1e-6;
const int mod = 1e9 + 7;
const int inf = 1e8;
const int maxn = 1e5 + 5;
char s[maxn];
int sta[maxn], cnt[maxn];
int main() {
  scanf("%s", s + 1);
  int tp = 0, ans = 0, L, R;
  sta[tp++] = 0;
  for (int i = 1; s[i]; i++) {
    cnt[i] = cnt[i - 1] + (s[i] == '[');
    if (s[i] == '(' || s[i] == '[') {
      sta[tp++] = i;
    } else if ((s[i] == ')' && s[sta[tp - 1]] != '(') ||
               (s[i] == ']' && s[sta[tp - 1]] != '[')) {
      tp = 0;
      sta[tp++] = i;
    } else {
      tp--;
      if (cnt[i] - cnt[sta[tp - 1]] > ans) {
        ans = cnt[i] - cnt[sta[tp - 1]];
        L = sta[tp - 1];
        R = i;
      }
    }
  }
  cout << ans << endl;
  if (ans) {
    for (int i = L + 1; i <= R; i++) {
      printf("%c", s[i]);
    }
  }
}
