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;
void openfile() {}
const int INF = 1e9;
const int N = 1e6 + 22;
char s[N];
int q[N], cnt[N];
char le[222];
int main() {
  int n, i, j;
  openfile();
  while (scanf("%s", s + 1) != EOF) {
    n = strlen(s + 1);
    le[')'] = '(';
    le[']'] = '[';
    int top = 0, ans = 0, L = 0, R = 0;
    for (i = 1; i <= n; ++i) {
      cnt[i] = cnt[i - 1] + (s[i] == '[');
      if (s[i] == '(' || s[i] == '[')
        q[top++] = i;
      else {
        if (top && s[q[top - 1]] == le[s[i]]) {
          top--;
          if (cnt[i] - cnt[q[top - 1]] > ans)
            ans = cnt[i] - cnt[q[top - 1]], L = q[top - 1], R = i;
        } else {
          q[0] = i;
          top = 1;
        }
      }
    }
    cout << ans << endl;
    for (i = L + 1; i <= R; ++i) putchar(s[i]);
    puts("");
  }
}
