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 long long MOD = 1e9 + 7;
const double err = 1e-12;
const double PI = 3.141592653589793;
const int N = 1e5 + 5;
int mx = 0;
string s;
stack<int> st;
vector<int> pre(N, 0);
void solve() {
  int lid = 0, rid = -1;
  cin >> s;
  st.push(0);
  for (int i = 0; i < s.length(); i++) {
    pre[i + 1] = pre[i] + (s[i] == '[');
    if (s[i] == '[' || s[i] == '(')
      st.push(i + 1);
    else {
      if (st.size() != 1 && ((s[i] == ')' && s[st.top() - 1] == '(') ||
                             (s[i] == ']' && s[st.top() - 1] == '['))) {
        st.pop();
        if (pre[i + 1] - pre[st.top()] > mx) {
          mx = pre[i + 1] - pre[st.top()];
          lid = st.top();
          rid = i;
        }
      } else
        st.push(i + 1);
    }
  }
  cout << mx << "\n" << s.substr(lid, rid - lid + 1) << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}
