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 double EPS = 1e-9;
const int i32INF = 1 << 30;
const long long i64INF = 1LL << 60;
const double mINF = 1e100;
class item {
 public:
  item(int _from = 0, int _to = 0, int _v = 0) : from(_from), to(_to), v(_v) {}
  int from, to, v;
};
char s[100100];
int sk[100100];
int to[100100];
int getSk(int i) {
  if (i < 0) return 0;
  return sk[i];
}
int main() {
  gets(s);
  int n = strlen(s);
  for (int i = 0; i < n; i++)
    if (s[i] == '[') sk[i] = 1;
  for (int i = 1; i < n; i++) sk[i] += sk[i - 1];
  if (n == 1) {
    cout << 0 << endl;
    return 0;
  }
  stack<pair<char, int> > st;
  for (int i = 0; i < n; i++) {
    if (s[i] == ']') {
      if (!st.empty() && st.top().first == '[') {
        to[st.top().second] = i;
        st.pop();
      } else
        st = stack<pair<char, int> >();
    } else if (s[i] == ')') {
      if (!st.empty() && st.top().first == '(') {
        to[st.top().second] = i;
        st.pop();
      } else
        st = stack<pair<char, int> >();
    } else {
      st.push(make_pair(s[i], i));
    }
  }
  vector<item> v;
  for (int i = 0; i < n; i++)
    if (to[i]) v.push_back(item(i, to[i], getSk(to[i]) - getSk(i - 1)));
  if (v.empty()) {
    cout << 0 << endl;
    return 0;
  }
  vector<item> r(1, v[0]);
  for (int i = 1; i < v.size(); i++) {
    if (r.back().to == v[i].from - 1) {
      r.back().to = v[i].to;
      r.back().v += v[i].v;
    } else if (r.back().to < v[i].from - 1)
      r.push_back(v[i]);
  }
  item res(r[0]);
  for (int i = 0; i < r.size(); i++)
    if (r[i].v > res.v) res = r[i];
  s[res.to + 1] = 0;
  printf("%d\n%s\n", res.v, s + res.from);
  return 0;
}
