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;
string s;
int sum[Maxn];
int nil = 0;
stack<int> S;
int best = -1;
int l, r;
void Test(int cl, int cr) {
  if (cl > cr) return;
  if (sum[cr + 1] - sum[cl] > best) {
    best = sum[cr + 1] - sum[cl];
    l = cl;
    r = cr;
  }
}
int main() {
  std::ios_base::sync_with_stdio(0);
  getline(cin, s);
  sum[0] = 0;
  for (int i = 1; i <= s.length(); i++) sum[i] = sum[i - 1] + (s[i - 1] == '[');
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == ')')
      if (S.empty() || s[S.top()] != '(') {
        Test(S.empty() ? nil : S.top() + 1, i - 1);
        while (!S.empty()) S.pop();
        nil = i + 1;
      } else
        S.pop();
    else if (s[i] == ']')
      if (S.empty() || s[S.top()] != '[') {
        Test(S.empty() ? nil : S.top() + 1, i - 1);
        while (!S.empty()) S.pop();
        nil = i + 1;
      } else
        S.pop();
    else
      S.push(i);
    Test(S.empty() ? nil : S.top() + 1, i);
  }
  if (best == -1)
    cout << "0\n";
  else
    cout << best << endl << s.substr(l, r - l + 1) << endl;
  return 0;
}
