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 = 100005;
const int MOD = 1000000007;
vector<int> v;
string str;
void read() {
  cin >> str;
  str += "(";
}
void solve() {
  int n = str.size();
  for (int i = 0; i < n; i++) {
    if (str[i] == '(') v.push_back(i);
    if (str[i] == '[') v.push_back(i);
    if (str[i] == ')') {
      if (v.size() == 0 || str[v[v.size() - 1]] != '(')
        v.push_back(i);
      else
        v.pop_back();
    }
    if (str[i] == ']') {
      if (v.size() == 0 || str[v[v.size() - 1]] != '[')
        v.push_back(i);
      else
        v.pop_back();
    }
  }
  int m = 0, res = -1;
  for (int i = 0; i < v.size(); i++) {
    int fr = 0, to = 0;
    if (i == 0)
      fr = 0;
    else
      fr = v[i - 1] + 1;
    to = v[i];
    int cnt = 0;
    for (int j = fr; j < to; j++) cnt += (str[j] == '[');
    if (m <= cnt) {
      m = cnt;
      res = i;
    }
  }
  cout << m << '\n';
  if (res == -1)
    cout << '\n';
  else if (res == 0) {
    cout << str.substr(0, v[res]);
  } else {
    cout << str.substr(v[res - 1] + 1, v[res] - v[res - 1] - 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}
