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;
long long cnt[200009];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  memset(cnt, -1, sizeof cnt);
  long long i, j, k, l, m, n, a, b;
  long long mn = 0;
  string s;
  cin >> s;
  n = s.size();
  stack<long long> st;
  a = -1, b = -3;
  for (i = 0; i < n; i++) {
    if (s[i] == '(') st.push(i);
    if (s[i] == '[') st.push(i);
    if (s[i] == ')') {
      if (st.size() && s[st.top()] == '(') {
        cnt[st.top()] = i;
        st.pop();
      } else {
        while (st.size()) st.pop();
      }
    }
    if (s[i] == ']') {
      if (st.size() && s[st.top()] == '[') {
        cnt[st.top()] = i;
        st.pop();
      } else {
        while (st.size()) st.pop();
      }
    }
  }
  long long pok;
  for (i = n - 2; i >= 0; i--) {
    if (cnt[i] != -1 && cnt[i] + 1 < n) {
      cnt[i] = max(cnt[i], cnt[cnt[i] + 1]);
    }
  }
  for (i = 0; i < n;) {
    if (s[i] == ')' || s[i] == ']' || cnt[i] == -1) {
      i++;
      continue;
    }
    pok = 0;
    for (j = i; j <= cnt[i]; j++) {
      if (s[j] == '[') pok++;
    }
    if (pok > mn) {
      mn = pok;
      a = i;
      b = cnt[i];
    }
    i = j;
  }
  cout << mn << '\n';
  for (i = a; i <= b; i++) cout << s[i];
  cout << '\n';
}
