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 m = 1e6 + 5;
int ans[m];
vector<int> mach;
string s;
vector<int> u;
vector<int> p;
int main() {
  int ma = 0, sum = 0, max = 0;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(' or s[i] == '[') {
      mach.push_back(i);
    } else {
      if (mach.size() > 0) {
        if (s[mach.back()] == '[' && s[i] == ']') {
          ans[i + 1] = 1;
          ans[mach.back() + 1] = 1;
          mach.pop_back();
        } else if (s[mach.back()] == '(' && s[i] == ')') {
          ans[i + 1] = -1;
          ans[mach.back() + 1] = -1;
          mach.pop_back();
        } else
          mach.clear();
      }
    }
  }
  for (int j = 1; j <= s.size() + 1; j++) {
    if (ans[j] > 0) {
      sum++;
      u.push_back(j);
    } else if (ans[j] < 0) {
      u.push_back(j);
    } else if (ans[j] == 0) {
      if (sum > ma) {
        ma = sum;
        p.clear();
        for (int o = 0; o < u.size(); o++) {
          p.push_back(u[o]);
        }
      }
      u.clear();
      sum = 0;
    }
  }
  if (sum > ma) {
    p.clear();
    for (int o = 0; o < u.size(); o++) {
      p.push_back(u[o]);
    }
    ma = sum;
  }
  cout << ma / 2 << endl;
  if (ma != 0) {
    for (int y = p[0] - 1; y < p[p.size() - 1]; y++) {
      cout << s[y];
    }
  }
}
