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;
string s;
int n;
int maxLen;
int ofrom, oto;
int dfs(int depth, int &pos, char kind) {
  if (pos == n) return 0;
  int from, to;
  int curSum = 0;
  int now = pos;
  from = pos;
  while (1) {
    if (pos == n) {
      if (now == from) return curSum;
      return -1;
    }
    char ch = s[pos];
    if (ch == '(' || ch == '[') {
      pos++;
      char rev;
      if (ch == '(')
        rev = ')';
      else if (ch == '[')
        rev = ']';
      int a = dfs(depth + 1, pos, rev);
      if (a != -1) {
        curSum += a;
        to = pos - 1;
        if (maxLen < curSum) {
          maxLen = curSum;
          ofrom = from;
          oto = to;
        }
      } else {
        curSum = 0;
        from = pos;
      }
    } else {
      pos++;
      if (kind == ch) {
        if (from != now) return -1;
        if (ch == ']') return 1 + curSum;
        return curSum;
      }
      kind = 0;
      from = pos;
      curSum = 0;
      continue;
      return -1;
    }
  }
  return -1;
}
int main() {
  cin >> s;
  n = s.size();
  int pos = 0;
  dfs(0, pos, 0);
  cout << maxLen << endl;
  if (maxLen != 0) {
    for (int i = ofrom; i <= oto; i++) cout << s[i];
    cout << endl;
  }
  return 0;
}
