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 int MAXN = 2 * 100000;
bool a[MAXN];
int main() {
  int ans = 0, pos = 0;
  stack<pair<char, int> > st;
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(' || s[i] == '[') {
      st.push(make_pair((s[i] == '(' ? ')' : ']'), i));
      continue;
    }
    if (!st.empty() && s[i] == st.top().first) {
      a[st.top().second] = a[i] = 1;
      st.pop();
      continue;
    }
    while (!st.empty()) st.pop();
  }
  for (int i = 0; i < s.size(); i++) {
    int now = 0, nowPos = i;
    for (; i < s.size() && a[i]; i++)
      if (s[i] == '[') now++;
    if (now > ans) {
      ans = now;
      pos = nowPos;
    }
  }
  cout << ans << endl;
  for (int i = pos; i < s.size() && a[i]; i++) cout << s[i];
  return 0;
}
