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 int mod = 1000000007;
const long long int INF = 1e18;
const int N = 20;
map<long long int, long long int> m;
void solve(int t) {
  string s;
  cin >> s;
  vector<pair<long long int, long long int> > v;
  stack<pair<char, long long int> > st;
  long long int ma = 0;
  long long int n = s.length();
  long long int b[n];
  long long int cn = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '[' or s[i] == ']') cn++;
    b[i] = cn;
  }
  long long int x = 0;
  long long int y = -1;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '(' or s[i] == '[') {
      st.push({s[i], i});
    } else if (!st.empty()) {
      if (s[i] == ')' and st.top().first == '(') {
        st.pop();
        long long int p;
        if (st.empty())
          p = 0;
        else
          p = st.top().second + 1;
        long long int q = i;
        long long int num;
        if (p == 0) {
          num = b[q];
        } else
          num = b[q] - b[p - 1];
        if (num / 2 > ma) {
          ma = num / 2;
          x = p;
          y = q;
        }
      } else if (s[i] == ']' and st.top().first == '[') {
        st.pop();
        long long int p;
        if (st.empty())
          p = 0;
        else
          p = st.top().second + 1;
        long long int q = i;
        long long int num;
        if (p == 0) {
          num = b[q];
        } else
          num = b[q] - b[p - 1];
        if (num / 2 > ma) {
          ma = num / 2;
          x = p;
          y = q;
        }
      } else {
        st.push({s[i], i});
      }
    } else {
      st.push({s[i], i});
    }
  }
  cout << ma << "\n";
  cout << s.substr(x, y - x + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
}
