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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 1e5 + 5, M = 100 + 5, OO = 0x3f3f3f3f, mod = 1e9 + 7, base = 131,
          MOD2 = 1000136437;
string s, z, x;
int p[N];
stack<int> st;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  memset((p), (-1), sizeof(p));
  ;
  for (int i = 0; i < (int)(s).size(); i++) {
    if (s[i] == '(' || s[i] == '[')
      st.push(i);
    else {
      if (st.empty()) continue;
      char z = s[st.top()];
      if ((z == '(' && s[i] == ')') || (z == '[' && s[i] == ']'))
        p[i] = st.top(), p[st.top()] = i, st.pop();
      else
        while (!st.empty()) st.pop();
    }
  }
  int mxi = 0, j;
  for (int i = 0; i < (int)(s).size(); i++) {
    if (p[i] == -1) continue;
    int cur = 0;
    z = "";
    while (p[i] != -1) {
      for (j = i; j <= p[i]; j++) {
        cur += (s[j] == '[');
        z += s[j];
      }
      i = j;
    }
    if (cur > mxi) mxi = cur, x = z;
    i = j - 1;
  }
  cout << mxi << endl << x;
  return 0;
}
