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;
using ll = long long;
using ii = pair<int, int>;
using pll = pair<ll, ll>;
const int N = 100010;
char s[N];
int n, f[N], pref[N];
int cnt(int i, int j) { return pref[j] - pref[i - 1]; }
char opp(char a) {
  if (a == '[') return ']';
  return ')';
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  pref[0] = 0;
  for (int i = 1; i <= n; i++) pref[i] = pref[i - 1] + (s[i] == '[');
  memset(f, 0, sizeof f);
  for (int i = n; i >= 1; i--)
    if (s[i] == '(' || s[i] == '[') {
      int k = i + f[i + 1] + 1;
      if (s[k] != opp(s[i])) continue;
      f[i] = (k + 1 - i) + f[k + 1];
    }
  int mx = 0, ix = -1;
  for (int i = 1; i <= n; i++) {
    int t = cnt(i, i + f[i] - 1);
    if (t > mx) {
      mx = t;
      ix = i;
    }
  }
  printf("%d\n", mx);
  s[ix + f[ix]] = '\0';
  if (mx > 0) printf("%s\n", s + ix);
}
