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;
long long t, n, u, v, m, q, r, ql, qr, k, d, c, s, b, l;
const int N = 2e5 + 500;
const long long mod = 1e9 + 7;
const bool JUDGE = false;
int st[N];
int op[N];
string str;
int S[N];
int H = 1;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (JUDGE) {
    freopen("grader.txt", "r", stdin);
    freopen("grader0.txt", "w", stdout);
  }
  cin >> str;
  str = '$' + str;
  n = str.size() - 1;
  st[0] = 0;
  for (int i = 1; i <= n; ++i) {
    S[i] = S[i - 1] + (str[i] == '[');
    op[i] = -1;
  }
  for (int i = 1; i <= n; ++i) {
    if (str[i] == '(' || str[i] == '[') {
      st[H++] = i;
    } else if (str[i] == ')') {
      if (str[st[H - 1]] == '(') {
        op[i] = st[H - 1];
        H--;
      } else
        st[H++] = i;
    } else {
      if (str[st[H - 1]] == '[') {
        op[i] = st[H - 1];
        H--;
      } else
        st[H++] = i;
    }
  }
  int best = 0;
  int start = 0;
  int end = -1;
  for (int i = n; i >= 1; --i) {
    int x = i;
    if (op[i] == -1) continue;
    while (i >= 1 && op[i] != -1) {
      i = op[i] - 1;
    }
    if (i != x) i++;
    if (S[x] - S[i - 1] > best) {
      best = S[x] - S[i - 1];
      start = i;
      end = x;
    }
  }
  cout << best << '\n';
  for (int i = start; i <= end; ++i) cout << str[i];
  cout << '\n';
  return 0;
}
