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;
int n, i, j, sk[200000], ct[200000], ans, u, le;
char ss[200000], c;
void ue(int a, int u1, int l) {
  if (a > ans) ans = a, u = u1, le = l;
}
int main() {
  scanf("%s", &ss);
  n = strlen(ss);
  for (int i = (0); i <= (n - 1); i++) {
    if (i) ct[i] = ct[i - 1];
    if (ss[i] == '[') ct[i]++;
  }
  sk[0] = 0;
  for (int i = (0); i <= (n - 1); i++) {
    c = ss[i];
    if (sk[0]) {
      if (c == ')' && ss[sk[sk[0]]] == '(' ||
          c == ']' && ss[sk[sk[0]]] == '[') {
        sk[0]--;
        if (sk[0])
          ue(ct[i] - ct[sk[sk[0]]], sk[sk[0]] + 1, i - sk[sk[0]]);
        else
          ue(ct[i], 0, i + 1);
      } else {
        sk[++sk[0]] = i;
      }
    } else {
      sk[++sk[0]] = i;
    }
  }
  printf("%d\n", ans);
  if (ans)
    for (int i = (u); i <= (u + le - 1); i++) printf("%c", ss[i]);
  cout << endl;
}
